Change the Cartesian integral to an equivalent polar integral, and then evaluate.
-$\int _ { - 4 } ^ { 4 } \int _ { - \sqrt { 16 - y ^ { 2 } } } ^ { \sqrt { 16 - y ^ { 2 } } } \ln \left( x ^ { 2 } + y ^ { 2 } + 1 \right) d x d y$
A) $\pi ( 16 \ln 17 + 16 )$
B) $\pi ( 17 \ln 17$ - 16)
C) $\pi ( 16 \ln 17 - 16 )$
D) $\pi ( 17 \ln 17 + 16 )$

Question

Quizplus · Accepted Answer

The given Cartesian integral can be converted to polar coordinates where $x = r\cos(\theta)$ and $y = r\sin(\theta)$. The limits for $r$ are from 0 to 4 (the radius of the circle described by the limits of integration), and the limits for $\theta$ are from 0 to $2\pi$ (a full circle). The integral becomes $\int_0^{2\pi} \int_0^4 r \ln(r^2 + 1) dr d\theta$. Evaluating this integral gives $\pi ( 17 \ln 17 - 16 )$, which corresponds to choice B.

Change the Cartesian Integral to an Equivalent Polar Integral, and Then