Solve the problem.
-Find the average height of the paraboloid $z = 5 x ^ { 2 } + 10 y ^ { 2 }$ above the annular region $9 \leq x ^ { 2 } + y ^ { 2 } \leq 100$ in the $x y$-plane.
A) $\frac { 545 } { 2 }$
B) $\frac { 1635 } { 4 }$
C) $\frac { 2725 } { 4 }$
D) 545

Question

Quizplus · Accepted Answer

The average height is given by the integral of the function over the region divided by the area of the region. The area of the annular region is $\pi(100-9)=\pi(91)$. So the average height is$$\frac{1}{\pi(91)}\iint_R (5x^2+10y^2) dA=\frac{1}{\pi(91)}\int_0^{2\pi}\int_3^{10} (5r^2) r \ dr d\theta$$$$=\frac{1}{\pi(91)}\int_0^{2\pi}\left[\frac{5r^4}{4}\right]_3^{10} d\theta=\frac{1}{\pi(91)}\int_0^{2\pi}\frac{5(10^4-3^4)}{4} d\theta$$$$=\frac{1}{\pi(91)}\left[\frac{5(10^4-3^4)}{4}\theta\right]_0^{2\pi}=\frac{5(10^4-3^4)}{4\pi(91)}(2\pi)=\frac{5(10^4-3^4)}{91}=\frac{1635}{4}$$

Solve the Problem z=5x2+10y2z = 5 x ^ { 2 } + 10 y ^ { 2 }z=5x2+10y2

Solve the Problem $z = 5 x ^ { 2 } + 10 y ^ { 2 }$