Question

1. $iint_{r} sin(9x^{2}+4y^{2})da$, where $r$ is the region in the first quadrant bounded by the ellipse $9x^{2}+4y^{2}=1$ 29. $iint e^{x + y}da$ where $r$ is given by the inequality $|x|+|y|leq1$

Explanation:

Step1: Use change - of - variables for 28

Let $u = 3x$ and $v = 2y$. Then $x=\frac{u}{3}$, $y = \frac{v}{2}$, and the Jacobian $J=\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{6}$. The ellipse $9x^{2}+4y^{2}=1$ becomes $u^{2}+v^{2}=1$, and the region $R$ in the $xy$ - plane (first - quadrant part of the ellipse) corresponds to the first - quadrant part of the unit circle in the $uv$ - plane. The double - integral $\iint_{R}\sin(9x^{2}+4y^{2})dA$ becomes $\frac{1}{6}\iint_{S}\sin(u^{2}+v^{2})dudv$, where $S$ is the first - quadrant part of the unit circle. Converting to polar coordinates in the $uv$ - plane: $u = r\cos\theta$, $v = r\sin\theta$, $dudv = rdr d\theta$, and the limits of integration are $0\leq r\leq1$ and $0\leq\theta\leq\frac{\pi}{2}$. So the integral is $\frac{1}{6}\int_{0}^{\frac{\pi}{2}}\int_{0}^{1}\sin(r^{2})r drd\theta$.

Step2: Evaluate the inner integral

Let $t = r^{2}$, then $dt = 2r dr$. When $r = 0$, $t = 0$; when $r = 1$, $t = 1$. The inner integral $\int_{0}^{1}\sin(r^{2})r dr=\frac{1}{2}\int_{0}^{1}\sin(t)dt=\frac{1}{2}[-\cos(t)]_{0}^{1}=\frac{1}{2}(1 - \cos(1))$.

Step3: Evaluate the outer integral

The outer integral $\frac{1}{6}\int_{0}^{\frac{\pi}{2}}\frac{1}{2}(1 - \cos(1))d\theta=\frac{\pi}{24}(1 - \cos(1))$.

Step4: For 29, consider two cases based on the sign of $x$ and $y$

Since $|x|+|y|\leq1$, when $x\geq0,y\geq0$, the region is $x + y\leq1$ or $y\leq1 - x$ with $0\leq x\leq1$ and $0\leq y\leq1 - x$. When $x\geq0,y\lt0$, the region is $x - y\leq1$ or $y\geq x - 1$ with $0\leq x\leq1$ and $x - 1\leq y\leq0$. When $x\lt0,y\geq0$, the region is $-x + y\leq1$ or $y\leq x + 1$ with $- 1\leq x\leq0$ and $0\leq y\leq x + 1$. When $x\lt0,y\lt0$, the region is $-x - y\leq1$ or $y\geq - x - 1$ with $-1\leq x\leq0$ and $-x - 1\leq y\leq0$. But we can also use symmetry. The double - integral $\iint_{R}e^{x + y}dA$ can be written as $4\int_{0}^{1}\int_{0}^{1 - x}e^{x + y}dydx$.

Step5: Evaluate the inner integral for 29

$\int_{0}^{1 - x}e^{x + y}dy=[e^{x + y}]_{0}^{1 - x}=e - e^{x}$.

Step6: Evaluate the outer integral for 29

$\int_{0}^{1}(e - e^{x})dx=[ex - e^{x}]_{0}^{1}=e-(e - 1)=1$.

Answer:

For 28: $\frac{\pi}{24}(1 - \cos(1))$; for 29: $1$