section 15.3 double integrals in polar coordinates 1067\n13. ∬_d e^(-x² - y²) da, where d is the region…

section 15.3 double integrals in polar coordinates 1067\n13. ∬_d e^(-x² - y²) da, where d is the region bounded by the semicircle x = √(4 - y²) and the y - axis\n14. ∬_d cos √(x² + y²) da, where d is the disk with center the origin and radius 2\n15. ∬_r arctan(y/x) da, where r = {(x, y) | 1 ≤ x² + y² ≤ 4, 0 ≤ y ≤ x}\n16. ∬_d x da, where d is the region in the first quadrant that lies between the circles x² + y² = 4 and x² + y² = 2x

section 15.3 double integrals in polar coordinates 1067\n13. ∬_d e^(-x² - y²) da, where d is the region bounded by the semicircle x = √(4 - y²) and the y - axis\n14. ∬_d cos √(x² + y²) da, where d is the disk with center the origin and radius 2\n15. ∬_r arctan(y/x) da, where r = {(x, y) | 1 ≤ x² + y² ≤ 4, 0 ≤ y ≤ x}\n16. ∬_d x da, where d is the region in the first quadrant that lies between the circles x² + y² = 4 and x² + y² = 2x

Answer

Explanation:

Step1: Recall polar - coordinate transformation

In polar coordinates, $x = r\cos\theta$, $y = r\sin\theta$, and $dA=r\ dr\ d\theta$, and $x^{2}+y^{2}=r^{2}$.

Step2: Solve problem 13

The equation of the semi - circle $x=\sqrt{4 - y^{2}}$ can be written as $x^{2}+y^{2}=4$ with $x\geq0$. In polar coordinates, the region $D$ is described by $0\leq r\leq2$ and $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$. The integral $\iint_{D}e^{-x^{2}-y^{2}}dA$ becomes $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2}e^{-r^{2}}r\ dr\ d\theta$. Let $u = r^{2}$, then $du = 2r\ dr$. When $r = 0$, $u = 0$; when $r = 2$, $u = 4$. The integral is $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left[-\frac{1}{2}e^{-u}\right]{0}^{4}d\theta=\int{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{2}(1 - e^{-4})d\theta=\pi(1 - e^{-4})$.

Step3: Solve problem 14

The disk $D$ with center at the origin and radius $2$ is described by $0\leq r\leq2$ and $0\leq\theta\leq2\pi$. The integral $\iint_{D}\cos\sqrt{x^{2}+y^{2}}dA$ becomes $\int_{0}^{2\pi}\int_{0}^{2}\cos(r)r\ dr\ d\theta$. Use integration by parts. Let $u = r$, $dv=\cos(r)dr$, then $du = dr$, $v=\sin(r)$. $\int r\cos(r)dr=r\sin(r)+\cos(r)+C$. So $\int_{0}^{2\pi}\left[r\sin(r)+\cos(r)\right]{0}^{2}d\theta=\int{0}^{2\pi}(2\sin(2)+\cos(2)-1)d\theta = 2\pi(2\sin(2)+\cos(2)-1)$.

Step4: Solve problem 15

The region $R$ where $1\leq x^{2}+y^{2}\leq4$ and $0\leq y\leq x$ corresponds to $1\leq r\leq2$ and $0\leq\theta\leq\frac{\pi}{4}$ in polar coordinates. The integral $\iint_{R}\arctan(\frac{y}{x})dA$ becomes $\int_{0}^{\frac{\pi}{4}}\int_{1}^{2}\theta\cdot r\ dr\ d\theta$. First, integrate with respect to $r$: $\int_{0}^{\frac{\pi}{4}}\theta\left[\frac{1}{2}r^{2}\right]{1}^{2}d\theta=\frac{3}{2}\int{0}^{\frac{\pi}{4}}\theta\ d\theta$. Then integrate with respect to $\theta$: $\frac{3}{2}\left[\frac{1}{2}\theta^{2}\right]_{0}^{\frac{\pi}{4}}=\frac{3\pi^{2}}{64}$.

Step5: Solve problem 16

The circle $x^{2}+y^{2}=4$ has $r = 2$ in polar coordinates, and the circle $x^{2}+y^{2}=2x$ can be rewritten as $r^{2}=2r\cos\theta$, or $r = 2\cos\theta$. The region $D$ in the first - quadrant is described by $2\cos\theta\leq r\leq2$ and $0\leq\theta\leq\frac{\pi}{2}$. The integral $\iint_{D}x\ dA$ becomes $\int_{0}^{\frac{\pi}{2}}\int_{2\cos\theta}^{2}(r\cos\theta)r\ dr\ d\theta=\int_{0}^{\frac{\pi}{2}}\cos\theta\left[\frac{1}{3}r^{3}\right]{2\cos\theta}^{2}d\theta=\frac{8}{3}\int{0}^{\frac{\pi}{2}}\cos\theta(1 - \cos^{3}\theta)d\theta$. Let $u=\sin\theta$, then $du=\cos\theta d\theta$. $\frac{8}{3}\int_{0}^{1}(1 - (1 - u^{2})^{3/2})du$. Another way: $\frac{8}{3}\left[\sin\theta-\frac{1}{4}\sin\theta\cos^{3}\theta-\frac{3}{8}\sin\theta\cos\theta-\frac{3}{8}\theta\right]_{0}^{\frac{\pi}{2}}=\frac{8}{3}\left(1 - \frac{3\pi}{16}\right)=\frac{8}{3}-\frac{\pi}{2}$.

Answer:

  1. $\pi(1 - e^{-4})$
  2. $2\pi(2\sin(2)+\cos(2)-1)$
  3. $\frac{3\pi^{2}}{64}$
  4. $\frac{8}{3}-\frac{\pi}{2}$