Find the volume of the indicated region.
-the region bounded by the paraboloid $z = 16 - x ^ { 2 } - y ^ { 2 }$ and the $x y$-plane
A) $128 \pi$
B) $\frac { 128 } { 3 } \pi$
C) $64 \pi$
D) $\frac { 256 } { 3 } \pi$

Question

Quizplus · Accepted Answer

The volume of the region bounded by the paraboloid $z = 16 - x^2 - y^2$ and the $xy$-plane can be found using the method of cylindrical coordinates. The paraboloid intersects the $xy$-plane where $z = 0$, which gives $16 - x^2 - y^2 = 0$, or $x^2 + y^2 = 16$. This is a circle of radius 4 in the $xy$-plane. In cylindrical coordinates, $x^2 + y^2 = r^2$, so the equation of the paraboloid becomes $z = 16 - r^2$. The volume can be calculated by integrating the area of circular slices from $r = 0$ to $r = 4$. The area of a slice at radius $r$ is $\pi r^2$, and the thickness of a slice is $dz$, which corresponds to a change in height $z = 16 - r^2$. The volume integral in cylindrical coordinates is:$$V = \int_0^{2\pi} \int_0^4 (16 - r^2) r \, dr \, d\theta$$$$V = \int_0^{2\pi} d\theta \int_0^4 (16r - r^3) \, dr$$$$V = \int_0^{2\pi} \left[8r^2 - \frac{r^4}{4}\right]_0^4 \, d\theta$$$$V = \int_0^{2\pi} \left[8(16) - \frac{256}{4}\right] \, d\theta$$$$V = \int_0^{2\pi} 32 \, d\theta$$$$V = 32\theta |_0^{2\pi}$$$$V = 64\pi$$However, the correct calculation should be:$$V = \int_0^{2\pi} \int_0^4 (16 - r^2) r \, dr \, d\theta = \int_0^{2\pi} \left[\frac{16r^2}{2} - \frac{r^4}{4}\right]_0^4 \, d\theta$$$$V = \int_0^{2\pi} \left[32 - 64\right] \, d\theta = \int_0^{2\pi} 32 \, d\theta = 64\pi$$This calculation was incorrect in the final step. The correct calculation should yield:$$V = \int_0^{2\pi} \int_0^4 (16r - r^3) \, dr \, d\theta = \int_0^{2\pi} \left[8r^2 - \frac{r^4}{4}\right]_0^4 \, d\theta = \int_0^{2\pi} \left[128 - 64\right] \, d\theta = 64\pi$$The correct calculation actually yields:$$V = \int_0^{2\pi} \int_0^4 (16 - r^2) r \, dr \, d\theta$$$$V = 2\pi \int_0^4 (16r - r^3) \, dr$$$$V = 2\pi \left[\frac{16r^2}{2} - \frac{r^4}{4}\right]_0^4$$$$V = 2\pi \left[128 - 64\right]$$$$V = 128\pi$$Therefore, the correct volume is $128\pi$, which corresponds to choice A. My initial explanation contained a miscalculation.

Find the Volume of the Indicated Region z=16−x2−y2z = 16 - x ^ { 2 } - y ^ { 2 }z=16−x2−y2

Find the Volume of the Indicated Region $z = 16 - x ^ { 2 } - y ^ { 2 }$