Evaluate the iterated integral by converting to polar coordinates. Round the answer to two decimal places.
$$\int _ { - 5 } ^ { 5 } \int _ { 0 } ^ { \sqrt { 25 - y ^ { 2 } } } \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 / 2 } d x d y .$$
A) $1963.5$
B) $15.71$
C) $4908.74$
D) $39.27$
E) 625

Question

Quizplus · Accepted Answer

Explanation:Convert to polar coordinates using $x = r \cos \theta$ and $y = r \sin \theta$. Then, $dx dy = r dr d\theta$. The limits of integration become:$$-5 \le x \le 5 \Rightarrow -5 \le r \cos \theta \le 5 \Rightarrow -\pi/4 \le \theta \le \pi/4$$$$0 \le y \le \sqrt{25 - y^2} \Rightarrow 0 \le r \sin \theta \le \sqrt{25 - r^2 \sin^2 \theta} \Rightarrow 0 \le r \le 5$$So, the integral becomes:$$\int_{-\pi/4}^{\pi/4} \int_0^5 (r^2)^3 r dr d\theta = \int_{-\pi/4}^{\pi/4} \int_0^5 r^7 dr d\theta$$$$= \int_{-\pi/4}^{\pi/4} \left[\frac{r^8}{8}\right]_0^5 d\theta = \int_{-\pi/4}^{\pi/4} \frac{5^8}{8} d\theta$$$$= \left[\frac{5^8}{8} \theta\right]_{-\pi/4}^{\pi/4} = \frac{5^8}{8} \cdot \frac{\pi}{2} \approx 1963.5$$

Evaluate the Iterated Integral by Converting to Polar Coordinates ∫−55∫025−y2(x2+y2)3/2dxdy.\int _ { - 5 } ^ { 5 } \int _ { 0 } ^ { \sqrt { 25 - y ^ { 2 } } } \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 / 2 } d x d y .∫−55​∫025−y2​​(x2+y2)3/2dxdy.

Evaluate the Iterated Integral by Converting to Polar Coordinates $\int _ { - 5 } ^ { 5 } \int _ { 0 } ^ { \sqrt { 25 - y ^ { 2 } } } \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 / 2 } d x d y .$