Q. 9 - Calculus

Question

Finding geometric quantities with definite integrals: Set up and solve definite integrals to find each volume, surface area, or arc length that follows. Solve each volume problem both with disks/washers and with shells, if possible. 

The volume of the solid obtained by revolving the region between the graph of $f\left(x\right) = x^2$ and the $y-$axis for $0 \le x \le 2$ around (a) the $x$-axis and (b) the $y$-axis .

Step-by-Step Solution

Answer

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Step 1. Given Information.

The volume of the solid obtained by revolving the region between the graph of $f\left(x\right)=x^2$ and the $y-$axis for $0\le x\le 2$.

(a) Step 2. Calculation.

The volume is given by :

$V=2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{yg}\left(\mathrm{y}\right)\mathrm{dy}$

As $y=x^2$

$$\begin{gathered} V=2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{y}f\left(\mathrm{x}\right)dy \\ \mathrm{V}=2\mathrm{\pi }{\int }_0^{\sqrt{2}}\mathrm{y}·y^{2}dy \\ \mathrm{V}=2\mathrm{\pi }{\int }_0^{\sqrt{2}}{\mathrm{y}}^{3}dy \\ \mathrm{V}=2\mathrm{\pi }{\left[\frac{{\mathrm{y}}^4}{4}\right]}_0^{\sqrt{2}} \\ \mathrm{V}= \frac{2\mathrm{\pi }}{4}\times 4 \\ \mathrm{V}=2\mathrm{\pi } \end{gathered}$$

(b) Step 3. Around the y - axis.

The volume of solid revolving about the $y-$axis is

$$\begin{gathered} V=2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{xf}\left(\mathrm{x}\right)\mathrm{dx} \\ \mathrm{V}=2\mathrm{\pi }{\int }_0^2\mathrm{x}·{\mathrm{x}}^2\mathrm{dx} \\ \mathrm{V}=2\mathrm{\pi }{\int }_0^2{\mathrm{x}}^3\mathrm{dx} \\ \mathrm{V}=2\mathrm{\pi }{\left[\frac{{\mathrm{x}}^4}{4}\right]}_0^2 \\ \mathrm{V}= \frac{2\mathrm{\pi }}{4}\times 16 \\ \mathrm{V}=8\mathrm{\pi } \end{gathered}$$