Q. 23 - 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.    

$\mathrm{\pi }{\int }_1^5{\left(\frac{\mathrm{y}-1}{2}\right)}^2\mathrm{dy}$

Step-by-Step Solution

Answer

The required region is shown as,    

Step 1. Given Information

The given integral is $\mathrm{\pi }{\int }_1^5{\left(\frac{\mathrm{y}-1}{2}\right)}^2\mathrm{dy}$

Step 2. Explanation

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

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

$f\left(y\right)=\frac{y-1}{2} \mathrm{in} \mathrm{the} \mathrm{interval} [1,5]$

Express f(y) as x and rewrite the function,

$$\begin{gathered} x=\frac{y-1}{2} \\ 2x=y-1 \\ y=2x+1 \end{gathered}$$

The simplified function represents a straight line.

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