Q. 24 - 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.   

$\mathrm{\pi }{\int }_0^4(2^2-{\left(\sqrt{\mathrm{y}}\right)}^2)\mathrm{dy}$

Step-by-Step Solution

Answer

The required region is shown as,    

Step 1. Given Information

The given integral is $\mathrm{\pi }{\int }_0^4(2^2-{\left(\sqrt{\mathrm{y}}\right)}^2)\mathrm{dy}$

Step 2. Explanation

The given integral is of the form $\mathrm{\pi }{\int }_a^b\left(f{\left(y\right)}^2\right)\mathrm{dy}$ 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,

$f\left(y\right) =\sqrt{2^2-{\left(\sqrt{y}\right)}^2}=\sqrt{4-y} \mathrm{in} \mathrm{the} \mathrm{interval} [0,4]$

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

$$\begin{gathered} x=\sqrt{4-y} \\ 4-y=x^2 \\ y=4-x^2 \end{gathered}$$

The simplified function represents a parabola.

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