In-class work for 3 July 2012
1. Compute the integral by using polar coordinates
    a.) $\displaystyle\int\int_R \sqrt{4-x^2-y^2} dA$, where $R$ is the region lying above the $x$-axis inside the circle $x^2+y^2=9$.
    b.) $\displaystyle\int\int_R \arctan(y/x) dA$, where $R = \{ (x,y) | 1 \leq x^2+y^2 \leq 4, 0 \leq y \leq x \}$.
    c.) $\displaystyle\int_{-3}^3 \int_0^{\sqrt{9-x^2}} \sin(x^2+y^2) dy dx$

2. Electric charge is distributed over the disk $x^2+y^2 \leq 4$ so that the charge density at $9x,y)$ is $\sigma(x,y) = x+y+x^2+y^2$. Find the total charge on the disk.

3. Find the mass and center of mass of the lamina that occupies the region $D$ and has the given density function $\rho$.
    a.) $D = \{(x,y) | 0 \leq x \leq a, 0 \leq y \leq b \}$; $\rho(x,y) = cxy$ for a constant $c$
    b.) $D$ is the triangular region enclosed by the lines $x=0, y=x$ and $2x+y=6$; $\rho(x,y) = x^2$ 4. The boundary of a lamina consists of the semicircles $y=\sqrt{1-x^2}$ and $y=\sqrt{4-x^2}$ together with the portions of the $x$-axis that join them. Find the center of mass if the density at any point is inversely proportional to the distance from the origin.

5. Define a lamina by a region $D$ that is bounded by $y=e^x$, $y=0, x=0, $ and $x=1$ with density function $\rho(x,y)=y$. Find the moments of inertia $I_x, I_y, I_0$ for this lamina.

6. A very mathematically important example! Consider the integral $\displaystyle\int_{-\infty}^{\infty} e^{-x^2} dx,$ which cannot be evaluated "normally". Notice that if we consider the similar looking integral $$\displaystyle\int\int_{\mathbb{R}^2} e^{-x^2-y^2} dA,$$ we can rewrite it as $\left( \displaystyle\int_{-\infty}^{\infty} e^{-x^2} dx \right) \left( \displaystyle\int_{-\infty}^{\infty} e^{-y^2} dy \right) = \left( \displaystyle\int_{-\infty}^{\infty} e^{-x^2} dx \right)^2,$ so that we can compue $\displaystyle\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\displaystyle\int\int_{\mathbb{R}^2} e^{-x^2-y^2} dA}$.

Use polar coordinates to evaluate $\displaystyle\int\int_{\mathbb{R}^2} e^{-x^2-y^2} dA$. What then does $\displaystyle\int_{-\infty}^{\infty} e^{-x^2} dx$ equal?