Q. 15 - Calculus

Question

Use the techniques of Section 13.2 to obtain an iterated integral that employs rectangular coordinates to represent the volume of the solid that is bounded above by the graph of the function $g$ from Exercise 14 and below by the $x-y$-plane over the circular disk $x^2+y^2\le b^2$.

Step-by-Step Solution

Answer

The volume of the solid generated is $V=\pi f\left(b\right)b^2$

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.

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

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

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

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

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

$V=2\pi f\left(b\right){\int }_0^{b}xdx$

$V=2\pi f\left(b\right){\left[\frac{x^2}{2}\right]}_0^b$

$V=\pi f\left(b\right)b^2$

Thus, the volume of the solid generated is

$V=\pi f\left(b\right)b^2$