Question

Use definite integrals to find the volume of each solid of revolution described in Exercises 49–61. (It is your choice whether to use disks/washers or shells in these exercises.)

The region between the graph of $f\left(x\right)=4-x^2$and the line y = 4 on [0, 2], revolved around the y-axis.

Short answer

The value of the volume is $8\pi$cubic units.

Step-by-step solution

Given information

We are given a function$f\left(x\right)=4-x^2$and y = 4

Find the integral and evaluate it

We know that the volume can be given as $V=2\pi {\int }_c^{d}r\left(x\right)h\left(x\right)dx$. The axis of revolution is y-axis

Hence the radius can be given as $r\left(x\right)=x$and the height is $h\left(x\right)=4-x^2$. Substituting the values in the integral we get,

$$\begin{gathered} V=2\pi {\int }_0^{2}x(4-x^2)dx \\ V=2\pi {\int }_0^{2}4x-x^{3}dx \\ V=2\pi {{[2x^2-\frac{x^4}{4}]}^2}_{0}V=2\pi \left[4\right] \\ V=8\pi \end{gathered}$$