How to find volume of a region - CLEP Calculus

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 2 and a major axes(the horizontal length) of 10, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 3 units to the left of the origin and the second cut is 2 units to the left of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or :

$V=\pi (2)^2(x-\frac{x^3}{3(5)^2})|_{-3}^{-2}$

$V=\pi (2)^2((-2)-\frac{(-2)^3}{3(5)^2})-\pi (2)^2((-3)-\frac{(-3)^3}{3(5)^2})$

$V=\frac{224}{75}\pi$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 1 and a major axes(the horizontal length) of 12, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 5 units to the left of the origin and the second cut is 4 units to the left of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or :

$V=\pi (x-\frac{x^3}{3(6)^2})|_{-5}^{-4}$

$V=\pi (-4-\frac{(-4)^3}{3(6)^2})-\pi (-5-\frac{(-5)^3}{3(6)^2})$

$V=\frac{47}{108}\pi$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 2 and a major axes(the horizontal length) of 6, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 2 units to the left of the origin and the second cut is 2 units to the right of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or :

$V=\pi (2)^2(x-\frac{x^3}{3(3)^2})|_{-2}^{2}$

$V=\pi (2)^2(2-\frac{2^3}{3(3)^2})-\pi (2)^2(-2-\frac{(-2)^3}{3(3)^2})$

$V=\frac{368}{27}\pi$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 1 and a major axes(the horizontal length) of 50, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 2 units to the left of the origin and the second cut is 14 units to the right of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or :

$V=\pi (x-\frac{x^3}{3(25)^2})|_{-2}^{14}$

$V=\pi (14-\frac{14^3}{3(25)^2})-\pi (-2-\frac{(-2)^3}{3(25)^2})$

$V=\frac{27248}{1875}\pi$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 8 and a major axes(the horizontal length) of 20, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is at the origin and the second cut is 5 units to the right of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or :

$V=\pi (8)^2(x-\frac{x^3}{3(10)^2})|_{0}^{5}$

$V=\pi (8)^2(5-\frac{5^3}{3(10)^2})-\pi (8)^2(0-\frac{0^3}{3(10)^2})$

$V=\frac{880}{3}\pi$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 2 and a major axes(the horizontal length) of 5, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 2 units to the left of the origin and the second cut is 1 unit to the left of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or $a=\frac{5}{2}$ :

$V=\pi (2)^2(x-\frac{x^3}{3(\frac{5}{2})^2})|_{-2}^{-1}$

$V=\pi (2)^2(-1-\frac{(-1)^3}{3(\frac{5}{2})^2})-\pi (2)^2(-2-\frac{(-2)^3}{3(\frac{5}{2})^2})$

$V=\frac{188}{75}\pi$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 1 and a major axes (the horizontal length) of 3, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 1 unit to the left of the origin and the second cut is at the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or $a=\frac{3}{2}$ :

$V=\pi (x-\frac{x^3}{3(\frac{3}{2})^2})|_{-1}^{0}$

$V=\pi (0-\frac{0^3}{3(\frac{3}{2})^2})-\pi (-1-\frac{(-1)^3}{3(\frac{3}{2})^2})$

$V=\frac{23}{27}\pi$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of $\frac{1}{4}$ and a major axes (the horizontal length) of, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is $\frac{1}{10}$ units to the left of the origin and the second cut is $\frac{1}{10}$ units to the right of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius $b=\frac{1}{4}$ . and the length of the spheroid , or $a=\frac{1}{2}$ :

$V=\pi (\frac{1}{4})^2(x-\frac{x^3}{3(\frac{1}{2})^2})|_{-\frac{1}{10}}^{\frac{1}{10}}$

$V=\pi (\frac{1}{4})^2((\frac{1}{10})-\frac{(\frac{1}{10})^3}{3(\frac{1}{2})^2})-\pi (\frac{1}{4})^2((-\frac{1}{10})-\frac{(-\frac{1}{10})^3}{3(\frac{1}{2})^2})$

$V=\frac{37}{3000}\pi$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 0.8 and a major axes (the horizontal length) of 2, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 0.6 units to the left of the origin and the second cut is 0.4 units to the left of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or :

$V=\pi (0.8)^2(x-\frac{x^3}{3})|_{-0.6}^{-0.4}$

$V=\pi (0.8)^2(-0.4-\frac{(-0.4)^3}{3})-\pi (0.8)^2(-0.6-\frac{(-0.6)^3}{3})$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 0.3 and a major axes (the horizontal length) of 0.9 , has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 0.3 units to the left of the origin and the second cut is 0.1 units to the left of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or :

$V=\pi (0.3)^2(x-\frac{x^3}{3(0.45)^2})|_{-0.3}^{-0.1}$

$V=\pi (0.3)^2(-0.1-\frac{(-0.1)^3}{3(0.45)^2})-\pi (0.3)^2(-0.3-\frac{(-0.3)^3}{3(0.45)^2})$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 3 and a major axes (the horizontal length) of 8, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 4 units to the left of the origin and the second cut is at the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or :

$V=\pi (3)^2(x-\frac{x^3}{3(4)^2})|_{-4}^{0}$

$V=\pi (3)^2(0-\frac{0^3}{3(4)^2})-\pi (3)^2(-4-\frac{(-4)^3}{3(4)^2})$

$V=24\pi$

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Question

A prolate spheroid (two out of three axes are equal, and less than the third) with an equaitorial radius (the length of the two equal axes) of 6 and a major axes (the horizontal length) of 14, has a segment with perpendicular planes cut out of it, such as is pictured above.

If the first cut is 7 units to the left of the origin and the second cut is 7 units to the right of the origin, what is the volume of the segment?

Answer

To approach this problem, begin by picturing how the spheroid might be projected onto the two-dimensional Cartesian plane:

In its two-dimensional projection, we notice the outline of an ellipse. The formula for an ellipse with horizontal and vertical axes, and in the Cartesian coordinate system is given as:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Which can be rewritten in terms of as $y=b\sqrt{1-\frac{x^2}{a^2}}$

Now note the disk method of volume creation wherein we rotate a function around an axis (for instance, the x-axis):

$V=\int_{m}^{n}\pi f^2(x)dx$

The new function in the integral is akin to the formula of the volume of a cylinder:

$\pi hr_{cylinder}^2$ where $r_{cylinder}^2\rightarrow f^2(x)$ and $h \rightarrow dx$

The integral sums up these thin cylinders to give the volume of the shape.

Treating as our , this integral can be written as:

$V=\int_{m}^{n}\pi b^2(1-\frac{x^2}{a^2})dx$

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{m}^{n}$

Consider the points where the ellipse contacts the x-axis, the greatest and smallest possible values of on the ellipse:

$V=\pi b^2(x-\frac{x^3}{3a^2})|_{-a}^{a}$

$V=\pi b^2(a-\frac{a^3}{3a^2})-\pi b^2(-a-\frac{(-a)^3}{3a^2})$

$V=\pi b^2(\frac{2}{3}a)-\pi b^2(-\frac{2}{3}a)$

$V=\frac{4}{3}\pi ab^2$

This is how the volume of a prolate/oblate spheroid can be derived, and similary, this is how we can derive the area of our segment. We're told that our equatorial radius . and the length of the spheroid , or :

$V=\pi (6)^2(x-\frac{x^3}{3(7)^2})|_{-7}^{7}$

$V=\pi (6)^2(7-\frac{7^3}{3(7)^2})-\pi (6)^2(-7-\frac{(-7)^3}{3(7)^2})$

$V=336\pi$

In the case of how this problem was defined, it turns out we were asked to find the area of the full spheroid.

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Question

Suppose I want to construct a cylindrical container. It costs 5 dollars per square foot to construct the two circular ends and 2 dollars per square foot for the rounded side. If I have a budget of 100 dollars, what's the maximum volume possible for this container?

Answer

Write down equations representing the volume(V) and cost of the cylinder.

$V=\pi r^2h$

$100=5(2\pi r^2)+2(2\pi rh)=10\pi r^2 +4\pi rh$

We want to find the values of and that will maximize the volume. Before taking the derivative of the volume equation, let's eliminate by using the cost equation.

$h=\frac{100-10\pi r^2}{4\pi r}$

Plug this into the volume equation

$V=\pi r^2\left (\frac{100-10\pi r^2}{4\pi r} \right )=r\left (\frac{100-10\pi r^2}{4} \right )$

$V=\frac{100r-10\pi r^3}{4}$

So now is eliminated in the volume equation. Take the derivative with respect to , set it equal to zero, and solve for .

$\frac{dV}{dr}=\frac{100-30\pi r^2}{4}=0$

$r=\sqrt{\frac{10}{3\pi}}$

We can use this value to find

$h=\frac{100-10\pi \left (\sqrt{\frac{10}{3\pi}} \right )^2}{4\pi \sqrt{\frac{10}{3\pi}}}=\frac{100-\frac{100}{3}}{4\sqrt{\frac{10\pi}{3}}}=\sqrt{\frac{3}{10\pi}} \frac{200}{12}$

$h=\frac{50}{3}\sqrt{\frac{3}{10\pi}}$

Now that we found AND that maximizes the volume, we can find that maximum volume

$V=\pi r^2h=\pi\left (\sqrt{\frac{10}{3\pi}} \right )^2 \frac{50}{3}\sqrt{\frac{3}{10\pi}}$

$V=\frac{50}{3}\sqrt{\frac{10}{3\pi}}$

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Question

We have the function $f(x)=\sqrt{x}$ and it is used to form a three dimensional figure by rotating it about the line $y=4$ . Find the volume of that figure from $x=0$ to $x=5$ .

Answer

Imagine a test rectangle with a length of $4-\sqrt{x}$ and it is rotated around $y=4$ to form a cicular disk with area $\pi (4-\sqrt{x})^2$ . The disk has a thickness $\Delta x$ so that its volume is $\pi(4-\sqrt{x})^2 \Delta x$ . To find the total volume of the figure, turn this into an integral.

$Volume=\pi\int_{0}^{5}(4-\sqrt{x})^2 dx$

Perform the integration

$Volume=\pi\int_{0}^{5}(x-8\sqrt{x}+16) dx=\pi\left ( \frac{1}{2}(5)^2-\frac{16}{3}(5)^{\frac{3}{2}}+16(5) \right )$

$Volume=103.269$

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Question

Consider a volume V of height H along some axis, which we will simply call h, such that . We can express the figure as cross-sectional areas A(h) perpendicular to this h-axis. For example for a cone whose base has radius r, we can choose the axis to go through the point of the cone at h = 0 and then the cone is a stack of little circles of radius , so $A(h) = \pi r^2 h^2/H^2$ .

What is in terms of ?

Answer

The volume is approximated by the Riemann sum formed when we stack n layers atop each other with $\Delta h = H/n$ ,

$V \approx \sum_{k=1}^n A(h_k) ~ \Delta h$

In the limit as n gets large, this expression becomes the integral:

$V = \int_0^H A(h) ~dh$

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Question

What is the volume inside the bowl , ? Hint: This is a solid of revolution about the z-axis with a radius of $\sqrt{x^2 + y^2}$ .

Answer

Since we know that the volume element

$dV = A ~ dz = \pi ~ r^2 ~ dz = \pi ~z ~dz$ ,

and integrating gives:

$V = \int_0^1 \pi ~z~ dz = \pi \left[ \frac{z^2}{2} \right ]_0^1 = \pi \left( \frac{1}{2} - 0 \right ) = \frac{\pi}{2}$

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Question

The following function:

Is rotated around the -axis to create a three-dimensional shape. What is the volume of this object within the interval to ?

Answer

Note that for a given value of x in the function:

The value of f(x) is the distance between the corresponding point on the curve and the x-axis. If the curve is rotated around the x-axis to create a three-dimensional object, f(x) can be seen as the radius of a circular cross-section of the object for any value of x.

To define the volume of this object, we can view it as a sum of infinitely thin disks stacked along the x-axis:

$V = \int \pi(f(x)^{2})dx$

Note how this follows the formula for the volume of a cylinder or disk

$( \pi \cdot height \cdot radius ^2 )$ ,

where is the radius and is the height.

Plugging in our function and range, we can rewrite this as:

$V = \int_{0}^{2} \pi(x+x^{2})^{2}dx$

or

$V = \int_{0}^{2}\pi(x^{4} + 2x^{3} + x^{2})$

Integrating this yields

$V = \pi \left(\frac{x^{5}}{5} + \frac{x^{4}}{2} + \frac{x^{3}}{3}\right)$

Using the upper bound of 2 and lower bound of 0 we find,

$V = \pi\left(\frac{2^{5}}{5}+\frac{2^{4}}{2} +\frac{2^{3}}{3}\right) - \pi\left(\frac{0^{5}}{5}+\frac{0^{4}}{2} +\frac{0^{3}}{3}\right) =\pi \frac{256}{15}$ .

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Question

Using the method of cylindrical disks, find the volume of the region of the graph of

revolved around the -axis on the interval .

Answer

The formula for the volume is given as

$V=\pi\int_{a}^{b}r^2 dx$

where .

As such,

$V=\pi\int_{0}^{10}(x^2)^2 dx=\pi\int_{0}^{10}x^4 dx$ .

When taking the integral, we will use the inverse power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Applying this rule we get

$\pi\int_{0}^{10}x^4 dx=\pi\left(\frac{1}{5}x^5\right)\left.\begin{matrix} , \end{matrix}\right|_{0}^{10}$ .

And by the corollary of the first Fundamental Theorem of Calculus,

$\pi\left(\frac{1}{5}x^5\right)\left.\begin{matrix} , \end{matrix}\right|_{0}^{10}=\pi \left(\left[\frac{1}{5}(10)^5\right]-\left[\frac{1}{5}(0)^5\right]\right)=20000\pi$ .

As such, the volume is

$20000\pi$ units cubed.

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Question

Using the method of cylindrical disks, find the volume of the region of the graph of

$f(x)=\sqrt{x}$

revolved around the -axis on the interval .

Answer

The formula for the volume is given as

$V=\pi\int_{a}^{b}r^2 dx$

where .

As such,

$V=\pi\int_{0}^{3}(\sqrt{x})^2 dx$ .

When taking the integral, we will use the inverse power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Applying this rule we get

$V=\pi\int_{0}^{3}(\sqrt{x})^2 dx=\pi\int_{0}^{3}x, dx=\pi\left(\frac{1}{2}x^2\right)\left.\begin{matrix} , \end{matrix}\right|_{0}^{3}$ .

And by the corollary of the first Fundamental Theorem of Calculus,

$\pi\left(\frac{1}{2}x^2\right)\left.\begin{matrix} , \end{matrix}\right|_{0}^{3}= \pi\left(\left[\frac{1}{2}(3)^2\right]-\left[\frac{1}{2}(0)^2\right]\right)$ .

As such, the volume is

$\frac{9}{2}\pi$ units cubed.

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Question

Find the volume of the equation $y=x^2 ; ; ; 0\leq x \leq 2$ revolved about the -axis.

Answer

This problem can be solved using the Disk Method and the equation

$\int_{a}^{b}\pi [y(x)]^2dx ; ; ; a\leq x\leq b$ .

Using our equation to formulate this equation we get the following.

$\ \int_{0}^{2}\ \pi x^4dx \ \$

Applying the power rule of integrals which states

$\int x^n=\frac{x^{n+1}}{n+1}$

we get,

$\ =\pi \frac{x^{4+1}}{4+1}\ \ =\frac{\pi x^5}{5}|_0^2\ \=\frac{32\pi }{5}$ .

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