Change to Polar Coordinates:

Evaluate the given integral by changing to polar coordinates:

• $\iint_R (2x-y) dA$, where $R$ is the region in the first quadrant enclosed by the circle $x^2+y^2 = 4$ and the lines $x = 0$ and $y = x$.
• $\iint_R \frac{y^2}{x^2+y^2} dA$, where $R$ is the region that lies between the circles $x^2+y^2 = a^2$ and $x^2 + y^2 = b^2$ with $0 \lt a \lt b$.
• $\int_{-3}^3 \int_0^{\sqrt{9-x^2}} \sin(x^2+y^2) dy dx$
• $\int_0^2 \int_0^{\sqrt{2x-x^2}} \sqrt{x^2+y^2} dy dx$

Area/Volume

Use a double integral to find the area of the region enclosed by both of the cardioids $r = 1 + \cos \theta$ and $ r = 1 - \cos \theta$.

Use polar coordinates to find the volume of the given solid:

• Under the cone $z = \sqrt{x^2+y^2}$ and above the disk $x^2 + y^2 \leq 4$.
• Bounded by the paraboloids $z = 3x^2 + 3y^2$ and $z = 4 -x^2 -y^2$.

Interesting:

Let $D$ be the disk with center the origin and radius $a$. What is the average distance from points in $D$ to the origin?

Use polar coordinates to combine the sum $$\int_{1/\sqrt{2}}^1 \int_{\sqrt{1-x^2}}^x xy dy dx + \int_1^{\sqrt{2}} \int_0^x xy dy dx + \int_{\sqrt{2}}^2 \int_0^{\sqrt{4-x^2}} xy dy dx$$ into one double integral. Then evaluate the double integral.