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

Converting to polar coordinates gives$$\begin{split} x & = r \cos 	heta \\ y & = r \sin 	heta \\ dx dy & = r \ dr \ d	heta \end{split}$$and the region of integration becomes$$0 \le r \le 5, \quad 0 \le 	heta \le \pi/2.$$So the integral becomes$$\begin{split} \int_0^{\pi/2} \int_0^5 (r^2)^3 r \ dr \ d	heta & = \int_0^{\pi/2} \int_0^5 r^7 \ dr \ d	heta \\ & = \int_0^{\pi/2} \left[ \frac{1}{8}r^8 ight]_0^5 \ d	heta \\ & = \int_0^{\pi/2} \frac{390625}{8} \ d	heta \\ & = \left[ \frac{390625}{8} 	heta ight]_0^{\pi/2} \\ & = \frac{390625\pi}{16} \approx 1963.49 \end{split}$$

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 $\int _ { - 5 } ^ { 5 } \int _ { 0 } ^ { \sqrt { 25 - y ^ { 2 } } } \left( x ^ { 2 } + y ^ { 2 } \right) ^ { 3 / 2 } d x d y .$