Set up and evaluate a double integral to find the volume of the solid bounded by the graphs of the equations given below.

Question

Quizplus · Accepted Answer

Since we are finding the volume of the solid, we need to set up a double integral in terms of $x, y,$ and $z.$
The solid is bounded by the graphs of $z = x+y,$ $z = 4-x,$ $x = 0,$ and $y = 0.$
Note that the solid is bounded by planes, so a good idea is to integrate with respect to $z$ first. The bounds for $z$ are $x+y \leq z \leq 4-x.$
The projection of the solid onto the $xy$-plane is a triangle with vertices $(0,0), (4,0),$ and $(0,4).$ Thus the bounds for $x$ and $y$ are $0 \leq x \leq 4-y$ and $0 \leq y \leq 4.$

Therefore, the integral is:

\begin{align*}
\int_{0}^{4} \int_{0}^{4-y} \int_{x+y}^{4-x} \,dz\,dx\,dy &= \int_{0}^{4} \int_{0}^{4-y} \left[(4-x)-(x+y)ight]\,dx\,dy \
&= \int_{0}^{4} \int_{0}^{4-y} (4-2x-y)\,dx\,dy\
&= \int_{0}^{4} \left[4x-x^2-xyight]_{0}^{4-y}\,dy\
&= \int_{0}^{4} (8y-y^2-16)\,dy\
&= \left[4y^2 - \frac{1}{3} y^3 - 16yight]_{0}^{4}\
&= \boxed{\frac{32}{3}}
\end{align*} We need to find the volume of the solid bounded by the graphs of the equations:

$z = x^2 + y^2$ and $z = 8 - x^2 - y^2$

We can see that these are both equations of circular paraboloids. The first one opens upwards and the second one opens downwards. The region of intersection of these two paraboloids lies in the plane $z=4$.

To set up the double integral, we need to integrate over the region in the $xy$ plane that is bounded by the circle $x^2 + y^2 = 4$. We can use polar coordinates for this region:

$0 \leq r \leq 2$
$0 \leq 	heta \leq 2\pi$

The limits for $z$ are from the equation of the lower paraboloid, $z=x^2+y^2$, to the equation of the upper paraboloid, $z=8-x^2-y^2$.

So our double integral becomes:

$\int_{0}^{2\pi} \int_{0}^{2} \int_{r^2}^{8-r^2} r\, dz\, dr\, d	heta$

Evaluating this integral gives the volume of the solid:

$\frac{32}{3}\pi$

Set Up and Evaluate a Double Integral to Find the Volume