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With Worked Solution![To determine the volume of a solid when a curve is notated about the x-axis the formula \\ \(V=\pi \displaystyle \int_a^b y^2\, d x\) is used\\ If the curve is rotated about the \(y\)-axis the formula \\ \(V=\pi \displaystyle \int_a^b x^2 \,d y \text { is used }\)\\ \begin{multicols}{2} \textbf{Example 1}\\ %question 30944 Find the volume enclosed by the surface generated when the curve \(x^2 + 3y^2 = 12\) is rotated about the \(x\)-axis. \\ \textbf{Example 1 solution}\\ Let \(y = 0\) to find the \(x\)-intercepts,\\ $\begin{aligned}x^2 = 12 &\Rightarrow x = \pm\sqrt{12}\\V &= \pi\displaystyle \int\, y^2 \,dy\\x^2 + 3y^2 &= 12\\\therefore y^2 &= \frac{12 - x^2}{3}\end{aligned}$\\ $\begin{aligned}V &= \pi\displaystyle \int_{-\sqrt{12}}^{\sqrt{12}} \, \frac{12 - x^2}{3}\,dx\\&= \frac{2\pi}{3}\displaystyle \int_0^{\sqrt{12}} \, 12 - x^2 \,dx \qquad \text{(even function)}\\&= \frac{2\pi}{3}\left[12x - \frac{x^3}{3}\right]_0^{\sqrt{12}}\\&= \frac{2\pi}{3}\left[12\sqrt{12} - \frac{12\sqrt{12}}{3}\right]\\&= \frac{32\pi\sqrt{3}}{3}\text{ units}^3\end{aligned}$\\ \columnbreak \textbf{Example 2}\\%question 12030 The region bounded by the parabola \(y = {x^2} - 2\) and the \(x\)-axis , is rotated about the \(y\)-axis. Find the volume of the solid generated. \\ \textbf{Example 2 solution}\\ $\begin{aligned} y=x^{2}-2 \rightarrow x^{2}=&y+2 \\ V=\pi \displaystyle \int_{-2}^{0} y+2 dy &=\pi\left[\frac{y^{2}}{2}+2 y\right]_{-2}^{0} \\ &=\pi[0-(2-4)] \\ &=2 \pi \mathrm{~u}^{3} \end{aligned}$\\ \end{multicols}](/media/05raxk33/volumes-of-solids-of-revolution.png)
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For Example: Part of the parabola \(y = {x^2}\) between \(x = 1\) and \(x = 2\) is rotated about the \(x\)-axis. Find the volume so formed.
\[\begin{align*}V = \pi \int\limits_1^2 {{{\left( {{x^2}} \right)}^2}} dx &= \pi \int\limits_1^2 {{x^4}} dx\\ &= \dfrac{\pi }{5}\left[ {{x^5}} \right]_1^2 = \dfrac{{31\pi }}{5}\,\,{u^3}\end{align*}\]
For Example: Part of the parabola \(y = {x^2}\) between \(x = 1\) and \(x = 2\) is rotated about the \(y\)-axis. Find the volume so formed.
\[\begin{align*}V &= \pi \int\limits_1^4 {y\,} dy\\ &= \dfrac{\pi }{2}\left[ {{y^2}} \right]_1^4 = \dfrac{{15\pi }}{2}\,\,{u^3}\end{align*}\]