Solve the problem.
-Evaluate
$$\int _ { - 4 } ^ { 4 } \int _ { - \sqrt { 16 - x ^ { 2 } } } ^ { \sqrt { 16 - x ^ { 2 } } } \int _ { x ^ { 2 } + y ^ { 2 } } ^ { 4 } d z d y d x$$
by transforming to cylindrical or spherical coordinates.
A) $16 \pi$
B) $\frac { 32 } { 3 } \pi$
C) $32 \pi$
D) $\frac { 64 } { 3 } \pi$

Question

Quizplus · Accepted Answer

The given integral is best evaluated by transforming to cylindrical coordinates due to the symmetry of the limits and the presence of $x^2 + y^2$. In cylindrical coordinates, $x^2 + y^2 = r^2$, $dzdydx$ becomes $rdzdrd\theta$, and the limits transform to cover a cylinder of radius 4 and height 4 (from $z = r^2$ to $z = 4$). The integral becomes $\int_{0}^{2\pi} \int_{0}^{4} \int_{r^2}^{4} r dz dr d\theta$, which evaluates to $\frac{64}{3}\pi$, indicating the volume of the described region.

Solve the Problem ∫−44∫−16−x216−x2∫x2+y24dzdydx\int _ { - 4 } ^ { 4 } \int _ { - \sqrt { 16 - x ^ { 2 } } } ^ { \sqrt { 16 - x ^ { 2 } } } \int _ { x ^ { 2 } + y ^ { 2 } } ^ { 4 } d z d y d x∫−44​∫−16−x2​16−x2​​∫x2+y24​dzdydx

Solve the Problem $\int _ { - 4 } ^ { 4 } \int _ { - \sqrt { 16 - x ^ { 2 } } } ^ { \sqrt { 16 - x ^ { 2 } } } \int _ { x ^ { 2 } + y ^ { 2 } } ^ { 4 } d z d y d x$