Question

Use the techniques of this section to obtain an iterated integral that employs polar coordinates to represent the volume of the solid discussed in Exercise 15.

Short answer

The volume of the solid generated is$V={\int }_0^{r-t}2\pi r{\cos }^2\theta f\left(r\right)dr$

Step-by-step solution

Given Information

The objective of this problem is to represents the volume of the solid of revolution obtained when the region bounded above by the graph of $f$and bounded below by $x$-axis on the interval $[0,b]$is rotated about the $z$-axis.

Calculation

Use shell method for the volume generated by the region bounded by the graph.

$V={\int }_0^{b}2\pi xzdx$

$V={\int }_0^{b}2\pi xg(x,y)dx$

$V={\int }_0^{b}2\pi xf\left(\sqrt{x^2+y^2}\right)dx$

Substitute $x=r\cos \theta ,y=r\sin \theta$

$V={\int }_0^{r-b}2\pi r\cos \theta f\left(\sqrt{r^2{\cos }^2\theta +r^2{\sin }^2\theta }\right)\cos \theta dr$

$V={\int }_0^{r-b}2\pi r{\cos }^2\theta f\left(r\right)dr$

Thus, the volume of the solid generated is

$V={\int }_0^{r-t}2\pi r{\cos }^2\theta f\left(r\right)dr$