Solve the problem.
-Find the volume of the solid bounded by the elliptical cone $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 81 } = \frac { z ^ { 2 } } { 16 }$ and the planes $z = 0$ and $z = 4$. (The area of an ellipse with semiaxes $a$ and $b$ is $\pi a b$.)
A) $24 \pi$ units $^ { 3 }$
B) $288 \pi$ units $^ { 3 }$
C) $96 \pi$ units $^ { 3 }$
D) 96 units $^ { 3 }$

Question

Quizplus · Accepted Answer

The volume of the solid can be found by integrating the area of the elliptical cross-sections from $z = 0$ to $z = 4$. The area of each cross-section is $\pi \cdot 8z \cdot \frac{9z}{4} = \frac{72\pi z^2}{4}$. Integrating this from 0 to 4 gives $\int_0^4 \frac{72\pi z^2}{4} \, dz = 96\pi$.

Solve the Problem x264+y281=z216\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 81 } = \frac { z ^ { 2 } } { 16 }64x2​+81y2​=16z2​

Solve the Problem $\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 81 } = \frac { z ^ { 2 } } { 16 }$