Combine the sum of the two iterated integrals into a single integral by converting to polar coordinates. Evaluate the resulting iterated integral. $$\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { x } \sqrt { x ^ { 2 } + y ^ { 2 } } d y d x + \int _ { 2 } ^ { 2 \sqrt { 2 } } \int _ { 0 } ^ { \sqrt { 8 - x ^ { 2 } } } \sqrt { x ^ { 2 } + y ^ { 2 } } d y d x$$
A) $\frac { 3 \sqrt { 2 } } { 8 \pi }$
B) $\frac { 3 \pi } { 8 \sqrt { 2 } }$
C) $\frac { 8 \pi } { 3 }$
D) $\frac { 4 } { 3 } \sqrt { 2 } \pi$
E) $\frac { 4 } { 3 } \pi$

Question

Combine the sum of the two iterated integrals into a single integral by converting to polar coordinates. Evaluate the resulting iterated integral. $$\int _ { 0 } ^ { 2 } \int _ { 0 } ^ { x } \sqrt { x ^ { 2 } + y ^ { 2 } } d y d x + \int _ { 2 } ^ { 2 \sqrt { 2 } } \int _ { 0 } ^ { \sqrt { 8 - x ^ { 2 } } } \sqrt { x ^ { 2 } + y ^ { 2 } } d y d x$$
A) $\frac { 3 \sqrt { 2 } } { 8 \pi }$
B) $\frac { 3 \pi } { 8 \sqrt { 2 } }$
C) $\frac { 8 \pi } { 3 }$ 
D) $\frac { 4 } { 3 } \sqrt { 2 } \pi$
E) $\frac { 4 } { 3 } \pi$

Quizplus · Accepted Answer

The correct answer is A because the integral can be evaluated as follows:$$\int_0^2 \int_0^x \sqrt{x^2 + y^2} dy dx + \int_2^{2\sqrt{2}} \int_0^{\sqrt{8 - x^2}} \sqrt{x^2 + y^2} dy dx$$$$= \int_0^{2\pi} \int_0^2 r dr d\theta$$$$= \int_0^{2\pi} \left[ \frac{r^2}{2} \right]_0^2 d\theta$$$$= \int_0^{2\pi} 2 d\theta$$$$= 4\pi$$Therefore, the value of the integral is $\frac{4\pi}{3}$.

Combine the Sum of the Two Iterated Integrals into a Single