Question

Use a double integral to find the volume of the solid bounded by the graphs of the equations. $$z=x^2,z=0,x=0,x=2,y=0,y=4$$

Short answer

The volume of the solid is $\frac{32}{3}$ cubic units.

Step-by-step solution

Understand the Geometrical Meaning

The equations $z=x^2,z=0,x=0,x=2,y=0,y=4$ define a solid that is essentially a 3D region under the curve $z=x^2$ between $x=0$ and $x=2$, and $y=0$ and $y=4$. The solid is bounded below by $z=0$ (the xy-plane).

Set Up Double Integral

The volume of a solid bounded by a surface $z=f(x,y)$ and plane $z=0$ can be calculated through the double integral ${\int }_Y{\int }_{X}f(x,y)dxdy$. Now, in this case, $f(x,y)=x^2$. Also, $0\le x\le 2$ and $0\le y\le 4$. So, the double integral becomes ${\int }_0^4{\int }_0^2{x}^{2}dxdy$.

Compute Inner Integral

The inner integral concentrates on the variable $x$. It can be computed by using standard integral rules, leading to ${\int }_0^2{x}^{2}dx={\left[\frac{1}{3}{x}^3\right]}_0^2=\frac{8}{3}$. The resulting (simplified) integral is ${\int }_0^4\frac{8}{3}dy$.

Compute Outer Integral

Now, let's compute the outer integral which is with respect to the variable $y$. This results in ${\int }_0^4\frac{8}{3}dy={\left[\frac{8}{3}y\right]}_0^4=\frac{32}{3}$.