Evaluate the iterated integral by converting to polar coordinates. Round the answer to two decimal places. $\int _ { - 3 } ^ { 3 } \int _ { 0 } ^ { \sqrt { 9 - y ^ { 2 } } } \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 / 2 } d x d y$ .
A) $152.68$
B) $5.65$
C) $14.14$
D) $48.6$
E) $381.7$

Question

Quizplus · Accepted Answer

Converting to polar coordinates gives $x = r \cos \theta$ and $y = r \sin \theta$, and $dx dy = r dr d\theta$. The limits of integration become $0 \le r \le 3$ and $0 \le \theta \le \pi/2$. So the integral becomes$$\int_0^{\pi/2} \int_0^3 (r^2)^{3/2} r \ dr d\theta = \int_0^{\pi/2} \left[\frac{1}{7}r^{7/2}\right]_0^3 d\theta = \int_0^{\pi/2} \frac{27}{7} d\theta = \frac{27\pi}{14} \approx 152.68.$$

Evaluate the Iterated Integral by Converting to Polar Coordinates ∫−33∫09−y2(x2+y2)3/2dxdy\int _ { - 3 } ^ { 3 } \int _ { 0 } ^ { \sqrt { 9 - y ^ { 2 } } } \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 / 2 } d x d y∫−33​∫09−y2​​(x2+y2)3/2dxdy

Evaluate the Iterated Integral by Converting to Polar Coordinates $\int _ { - 3 } ^ { 3 } \int _ { 0 } ^ { \sqrt { 9 - y ^ { 2 } } } \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 / 2 } d x d y$