# Kishan's question via email about volume by revolution

#### Prove It

##### Well-known member
MHB Math Helper
\displaystyle \begin{align*} R \end{align*} is the region in the first quadrant bounded above by the curve \displaystyle \begin{align*} y = 2 - 5\,\sqrt{x} \end{align*} and bounded below by \displaystyle \begin{align*} y = 0 \end{align*}. Find the volume of the solid obtained by rotating \displaystyle \begin{align*} R \end{align*} around the line \displaystyle \begin{align*} y = -4 \end{align*}.
We should first find the \displaystyle \begin{align*} x \end{align*} intercept of the function \displaystyle \begin{align*} y = 2 - 5\,\sqrt{x} \end{align*}, as this will be the end of our region of integration.

\displaystyle \begin{align*} 0 &= 2 - 5\,\sqrt{x} \\ 5\,\sqrt{x} &= 2 \\ \sqrt{x} &= \frac{2}{5} \\ x &= \frac{4}{25} \end{align*}

So our region of integration will be \displaystyle \begin{align*} 0 \leq x \leq \frac{4}{25} \end{align*}.

Then we need to realise that we have to find the volume of the solid obtained by rotating the entire region above \displaystyle \begin{align*} y = -4 \end{align*} and below \displaystyle \begin{align*} y = 2 - 5\,\sqrt{x} \end{align*} around the line \displaystyle \begin{align*} y = -4 \end{align*}, and then subtract the entire volume obtained by rotating the region between \displaystyle \begin{align*} y = -4 \end{align*} and \displaystyle \begin{align*} y = 0 \end{align*} around \displaystyle \begin{align*} y = -4 \end{align*}.

To do this, it will be easier if we move everything up four units, as then we will be rotating around the x axis (\displaystyle \begin{align*} y=0 \end{align*}) instead.

So we really want to find the volume obtained by rotating \displaystyle \begin{align*} y = 6 - 5\,\sqrt{x} \end{align*} around the x axis, and then subtracting the volume obtained by rotating \displaystyle \begin{align*} y = 4 \end{align*} around the x axis.

Now to find the volume of a solid obtained by rotating a function \displaystyle \begin{align*} y = f(x) \end{align*} around the x axis, we need to first picture the entire area and then picture it being rotated around the axis. We break up the area into rectangles, each of length \displaystyle \begin{align*} \Delta x \end{align*} (a small change in x) and of width \displaystyle \begin{align*} y = f(x) \end{align*}. Then if we rotate each rectangle, a cylinder is formed, which has a radius \displaystyle \begin{align*} y = f(x) \end{align*} and a height \displaystyle \begin{align*} \Delta x \end{align*}. Since the volume of a cylinder is \displaystyle \begin{align*} \pi \, r^2\,h \end{align*}, that means the volume of each cylinder is \displaystyle \begin{align*} \pi\,\left[ f(x) \right] ^2\,\Delta x \end{align*}.

The total volume can then be approximated by summing all these cylinder volumes, so \displaystyle \begin{align*} V \approx \sum{ \pi\,\left[ f(x) \right] ^2\,\Delta x } \end{align*}.

To improve on the approximation, we increase the number of rectangles. Each rectangle width gets smaller, and thus gives a better approximation of the volume. As \displaystyle \begin{align*} n \to \infty \end{align*} and \displaystyle \begin{align*} \Delta x \to 0 \end{align*}, the approximation becomes exact and the sum becomes an integral.

Therefore the volume of a solid formed by rotating \displaystyle \begin{align*} y = f(x) \end{align*} around the x axis is exactly \displaystyle \begin{align*} V = \int_a^b{ \pi\,\left[ f(x) \right] ^2\,\mathrm{d}x } \end{align*}.

So in the case of this example, we first have to find the volume of the solid formed by rotating \displaystyle \begin{align*} y = 6 - 5\,\sqrt{x} \end{align*} around the x axis, with \displaystyle \begin{align*} 0 \leq x \leq \frac{4}{25} \end{align*}, so this is

\displaystyle \begin{align*} V &= \int_0^{\frac{4}{25}}{ \pi\,\left( 6 - 5\,\sqrt{x} \right)^2 \,\mathrm{d}x } \\ &= \pi \int_0^{\frac{4}{25}}{ \left( 36 - 60\,\sqrt{x} + 25\,x \right) \,\mathrm{d}x } \\ &= \pi \int_0^{\frac{4}{25}}{ \left( 36 - 60\,x^{\frac{1}{2}} + 25\,x \right) \,\mathrm{d}x } \\ &= \pi \,\left[ 36\,x - \frac{60\,x^{\frac{3}{2}}}{\frac{3}{2}} + \frac{25\,x^2}{2} \right] _0^{\frac{4}{25}} \\ &= \pi\,\left[ 36\,x - 40\,x^{\frac{3}{2}} + \frac{25\,x^2}{2} \right]_0^{\frac{4}{25}} \\ &= \pi \,\left\{ \left[ 36\,\left( \frac{4}{25} \right) - 40\,\left( \frac{4}{25} \right) ^{\frac{3}{2}} + \frac{25\,\left( \frac{4}{25} \right) ^2}{2} \right] - \left[ 36\,\left( 0 \right) - 40\,\left( 0 \right) ^{\frac{3}{2}} + \frac{25\,\left( 0 \right) ^2 }{ 2} \right] \right\} \\ &= \pi \,\left[ \frac{144}{25} - 40 \, \left( \frac{8}{125} \right) + \frac{25 \,\left( \frac{16}{625} \right) }{2} - 0 \right] \\ &= \pi\,\left( \frac{144}{25} - \frac{64}{25} + \frac{8}{25} \right) \\ &= \frac{88\,\pi}{25} \end{align*}

We then need to subtract the volume formed by rotating the region under \displaystyle \begin{align*} y = 4 \end{align*} around the x axis. This is

\displaystyle \begin{align*} V &= \int_0^{\frac{4}{25}}{ \pi\,\left( 4 \right) ^2 \,\mathrm{d}x } \\ &= \pi\int_0^{\frac{4}{25}}{ 16\,\mathrm{d}x } \\ &= \pi\,\left[ 16\,x \right] _0^{\frac{4}{25}} \\ &= \pi\,\left[ 16\,\left( \frac{4}{25} \right) - 16\,\left( 0 \right) \right] \\ &= \frac{64\,\pi}{25} \end{align*}

Thus the volume we need is \displaystyle \begin{align*} \frac{88\,\pi}{25} - \frac{64\,\pi}{25} = \frac{24\,\pi}{25}\,\textrm{units}^3 \end{align*}.