Find the volume of the indicated region.
-the region that lies under the paraboloid $z = x ^ { 2 } + y ^ { 2 }$ and above the triangle enclosed by the lines $x = 8 , y = 0$, and $y = 2 x$
A) $\frac { 1792 } { 3 }$
B) 8192
C) $\frac { 14,336 } { 3 }$
D) $\frac { 224 } { 3 }$

Question

Quizplus · Accepted Answer

The triangle enclosed by the lines $x = 8 , y = 0$, and $y = 2 x$ has vertices at $(8,0)$, $(0,0)$, and $(4,8)$.

To find the limits of integration for the volume, we first need to find the limits of integration for $z$. The lowest point on the paraboloid is at the origin, so the minimum value of $z$ is 0. The maximum value of $z$ occurs at the point $(4,8)$, where $z=x^2+y^2=4^2+8^2=80$.

Next, we need to find the limits of integration for $x$ and $y$. We can do this by noting that the triangle is bounded on the left by the $y$ axis, on the right by the line $y=2x$, and on the bottom by the line $x=8$. Therefore, we have:

$$8 \leq x \leq 4y$$
$$0 \leq y \leq 8/4 = 2$$

So the volume of the region is given by:

$$\int_{0}^{2} \int_{8}^{4y} \int_{0}^{x^2+y^2} dz dx dy$$

Evaluating this triple integral gives:

$$\frac{14336}{3}$$

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

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