Solve the problem.
-Evaluate
$$\int _ { - 10 } ^ { 10 } \int _ { - \sqrt { 100 - x ^ { 2 } } } ^ { \sqrt { 100 - x ^ { 2 } } } \int _ { 0 } ^ { 6 } \left( x ^ { 2 } + y ^ { 2 } \right) z d z d y d x$$
by transforming to cylindrical or spherical coordinates.
A) $9000 \pi$
B) $90,000 \pi$
C) $120,000 \pi$
D) $900 \pi$

Question

Quizplus · Accepted Answer

The given integral is best evaluated by transforming to cylindrical coordinates due to the presence of $x^2 + y^2$, which simplifies to $r^2$ in cylindrical coordinates. The limits for $x$ and $y$ describe a circle with radius 10 centered at the origin, and $z$ ranges from 0 to 6. In cylindrical coordinates, $x^2 + y^2 = r^2$, $dz$ remains $dz$, and $dydx$ becomes $r dr d\theta$. The limits for $r$ are from 0 to 10, for $\theta$ are from 0 to $2\pi$, and for $z$ are from 0 to 6. The integral becomes:$$\int_{0}^{2\pi} \int_{0}^{10} \int_{0}^{6} r^2 z r dz dr d\theta = \int_{0}^{2\pi} d\theta \int_{0}^{10} r^3 dr \int_{0}^{6} z dz$$Evaluating the integrals gives:$$\int_{0}^{2\pi} d\theta = 2\pi$$$$\int_{0}^{10} r^3 dr = \left[\frac{1}{4}r^4\right]_0^{10} = \frac{1}{4}(10^4) = 2500$$$$\int_{0}^{6} z dz = \left[\frac{1}{2}z^2\right]_0^{6} = \frac{1}{2}(6^2) = 18$$Multiplying these results together gives:$$2\pi \times 2500 \times 18 = 90000\pi$$Therefore, the correct answer is $90,000 \pi$, which is choice B.

Solve the Problem By Transforming to Cylindrical or Spherical Coordinates 9000π9000 \pi9000π

Solve the Problem By Transforming to Cylindrical or Spherical Coordinates $9000 \pi$