Solve the problem.
-Find the volume of the solid bounded by the paraboloid $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 81 } = \frac { z } { 2 }$ and the planes $z = 0$ and $z = 6$. (The area of an ellipse with semiaxes $\mathrm { a }$ and $\mathrm { b }$ is $\pi \mathrm { ab }$.)
A) $567 \pi$ units $^ { 3 }$
B) 567 units $^ { 3 }$
C) $1134 \pi$ units $^ { 3 }$
D) $567 \pi ^ { 2 }$ units $^ { 3 }$

Question

Quizplus · Accepted Answer

The volume of the solid can be found by integrating the area of cross-sectional slices perpendicular to the z-axis from $z = 0$ to $z = 6$. The cross-section at a height $z$ is an ellipse with semiaxes $a = 7\sqrt{\frac{z}{2}}$ and $b = 9\sqrt{\frac{z}{2}}$, so its area is $\pi ab = \pi(7\sqrt{\frac{z}{2}})(9\sqrt{\frac{z}{2}}) = \frac{63\pi z}{2}$. Integrating this area from $z = 0$ to $z = 6$ gives the volume: $\int_{0}^{6} \frac{63\pi z}{2} dz = \frac{63\pi}{2} \left[\frac{z^2}{2}\right]_0^6 = \frac{63\pi}{2} \cdot 18 = 567\pi$ cubic units.

Solve the Problem x249+y281=z2\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 81 } = \frac { z } { 2 }49x2​+81y2​=2z​

Solve the Problem $\frac { x ^ { 2 } } { 49 } + \frac { y ^ { 2 } } { 81 } = \frac { z } { 2 }$