Q. 19 - Calculus

Question

Each of the definite integrals in Exercises 19–24 represents the volume of a solid of revolution obtained by rotating a region around either the x- or y-axis. Find this region.
$\mathrm{\pi }{\int }_1^3\left(16-{\left(\mathrm{x}+1\right)}^2\right)\mathrm{dx}$

Step-by-Step Solution

Answer

The required region is shown as,

Step 1. Given Information

The given integral is $\mathrm{\pi }{\int }_1^3(16-{(\mathrm{x}+1)}^2)\mathrm{dx}$

Step 2. Explanation

The given integral is of the form $\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}(\mathrm{f}{\left(\mathrm{x}\right))}^2\mathrm{dx}$ which represents the volume of a solid formed by rotating the region bound by f(x) and x-axis in the interval $[a,b]$ around x-axis.

On comparing the given region with the above region, we have,

$f\left(x\right)=\sqrt{16-{(x+1)}^2} \mathrm{in} \mathrm{the} \mathrm{interval} [1,3]$

The given function represents the semi circle with center $(-1,0)$ and radius 4 units.

Plot the graph of the function and identify the region between the intervals.