Q. 16 - Calculus

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.

Step-by-Step Solution

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 1 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.

Step 2 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

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