Calculus 3 - Double Integration in Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3cos(x^{2} + y^{2}))}{2}-\frac{ cos(2x^{2} + 2y^{2})}{8})dA\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.25\text{ and }1.93\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(-0.941,-0.339)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(6\cdot (\frac{1}{4})^{(x^{2} + y^{2})})}{19}-\frac{ (23e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{3})dA\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.27\text{ and }1.18\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-1.000,0.031)\text{ and }\overrightarrow{u_2}=(0.992,-0.125)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(47sin(x^{2} + y^{2}))}{2}-\frac{ e^{(x^{2} + y^{2})}}{4})dxdy\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.14\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.969,0.249)\text{ and }\overrightarrow{u_2}=(-0.988,0.156)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{23}- 32e^{(- 2x^{2} - 2y^{2})})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.26\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.969,0.249)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(12sin(2x^{2} + 2y^{2}) -\frac{ (5e^{(- 2x^{2} - 2y^{2})})}{41})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&0.9\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.827,0.562)\text{ and }\overrightarrow{u_2}=(0.661,-0.750)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(- 2e^{(- x^{2} - y^{2})} -\frac{ 40}{3})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.54\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.339,0.941)\text{ and }\overrightarrow{u_2}=(0.218,-0.976)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Evaluate the integral

$\int\int_D(x^2+y^2)dxdy$

where D is the region above the x-axis and within a circle centered at the origin of radius 2.

Evaluate the following integral by converting into Polar Coordinates.

$\int \int_D 10xy \ dA$ , where is the portion between the circles of radius and and lies in first quadrant.

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{33}{3^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}})dxdy\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.1\text{ and }1.76\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.707,0.707)\text{ and }\overrightarrow{u_2}=(0.063,-0.998)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(10cos(x^{2} + y^{2}))dxdy\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.49\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.063,0.998)\text{ and }\overrightarrow{u_2}=(-0.729,-0.685)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Double Integration in Polar Coordinates

Practice Questions