# Brenton's questions via email about volume by revolution

#### Prove It

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MHB Math Helper

Q5. Here is a graph of the region to be integrated and the line to be rotated around.

First we should find the x intercept of the function \displaystyle \begin{align*} y = 3 - 4\,\sqrt{x} \end{align*} as this will be the ending point of our region of integration.

\displaystyle \begin{align*} 0 &= 3 - 4\,\sqrt{x} \\ 4\,\sqrt{x} &= 3 \\ \sqrt{x} &= \frac{3}{4} \\ x &= \frac{9}{16} \end{align*}

To do this question, you need to first be able to visualise the solid. Picture the entire area above the line y = -2 and below \displaystyle \begin{align*} y = 3 - 4\,\sqrt{x} \end{align*} between x = 0 and \displaystyle \begin{align*} x = \frac{9}{16}\end{align*} being rotated around the line \displaystyle \begin{align*} y = -2 \end{align*}, and then picture the area between y = -2 and y = 0 over the same x region being rotated about the line y = -2 and removing it.

Notice that the resulting solid will be exactly the same if everything was moved up by 2 units, and so its volume will be exactly the same, with the advantage of being rotated around the x axis.

So we want to work out the volume of the region below \displaystyle \begin{align*} y = 5 - 4\,\sqrt{x} \end{align*} and above the x axis between \displaystyle \begin{align*} x = 0 \end{align*} and \displaystyle \begin{align*} x = \frac{9}{16} \end{align*}, and then subtract the volume of the region below \displaystyle \begin{align*} y = 2 \end{align*} and above the x axis in the same x region.

To evaluate these volumes, first picture the area as being cut up into a number of rectangles. When these rectangles are rotated, you get cylinders (horizontally oriented). So you can approximate the total volume by adding up the volumes of all the cylinders.

The volume of a cylinder is \displaystyle \begin{align*} \pi\,r^2\,h \end{align*}. In this case, the radius of each cylinder is the y value of the function, so \displaystyle \begin{align*} 5 - 4\,\sqrt{x} \end{align*}, and the height is a small change in x \displaystyle \begin{align*} \Delta x \end{align*}. That means the total volume can be approximated by \displaystyle \begin{align*} V \approx \sum{ \pi\,\left( 5 - 4\,\sqrt{x} \right) ^2\,\Delta x } \end{align*}.

To improve on the approximation, we increase the number of cylinders, making each cylinder thinner. 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. Thus the volume is exactly

\displaystyle \begin{align*} V &= \int_0^{\frac{9}{16}}{\pi\,\left( 5 - 4\,\sqrt{x} \right) ^2\,\mathrm{d}x } \\ &= \pi\int_0^{\frac{9}{16}}{ \left( 25 - 40\,\sqrt{x} + 16\,x \right) \,\mathrm{d}x } \\ &= \pi\int_0^{\frac{9}{16}}{ \left( 25 - 40\,x^{\frac{1}{2}} + 16\,x \right) \,\mathrm{d}x } \\ &= \pi\,\left[ 25\,x - \frac{40\,x^{\frac{3}{2}}}{\frac{3}{2}} + 8\,x^2 \right] _0^{\frac{9}{16}} \\ &= \pi\,\left[ 25\,x - \frac{80\,x^{\frac{3}{2}}}{3} + 8\,x^2 \right] _0^{\frac{9}{16}} \\ &= \pi\,\left\{ \left[ 25\,\left( \frac{9}{16} \right) - \frac{80\,\left( \frac{9}{16} \right) ^{\frac{3}{2}}}{3} + 8\,\left( \frac{9}{16} \right) ^2 \right] - \left[ 25\,\left( 0 \right) - \frac{80\,\left( 0 \right) ^{\frac{3}{2}}}{3} + 8\,\left( 0 \right) ^2 \right] \right\} \\ &= \pi \left\{ \left[ \frac{225}{16} - \frac{80\,\left( \frac{27}{64} \right) }{3} + 8\,\left( \frac{81}{256} \right) \right] - 0 \right\} \\ &= \pi \, \left( \frac{225}{16} - \frac{45}{4} + \frac{81}{32} \right) \\ &= \pi\,\left( \frac{450}{32} - \frac{360}{32} + \frac{81}{32} \right) \\ &= \frac{171\,\pi}{32}\,\textrm{units}^3 \end{align*}

Q6. Here is a graph of the region to be rotated about the y axis.

We need to first visualise this area being rotated around the y axis, to get a picture of what the solid would look like.

Then we need to imagine that this solid is being made up of thin vertically oriented hollow cylinders. We could then approximate the total volume by adding up the volumes of the cylinders.

Each cylinder's curved surface is a rectangle, with width equal to the y value of the function, and its length is the same as the circumference of the cylinder, so \displaystyle \begin{align*} 2\,\pi\,r \end{align*}. The radius of each cylinder is the x value of the function, so the area is \displaystyle \begin{align*} 2\,\pi\,x\,y \end{align*}. Thus the volume of each cylinder is \displaystyle \begin{align*} 2\,\pi\,x\,y\,\Delta x \end{align*}, where \displaystyle \begin{align*} \Delta x \end{align*} is a small change in x.

So the total volume is approximated by \displaystyle \begin{align*} V \approx \sum{ 2\,\pi\,x\,y\,\Delta x} \end{align*}.

If we increase the number of cylinders, making each cylinder thinner, we get a better approximation. So 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. So the volume is exactly

\displaystyle \begin{align*} V &= \int_1^{\frac{3}{2}}{ 2\,\pi\,x\,y\,\mathrm{d}x } \\ &= 2\,\pi\int_1^{\frac{3}{2}}{ x\,\left( 4 + \frac{1}{5}\,x^2 \right) \,\mathrm{d}x } \\ &= 2\,\pi\int_1^{\frac{3}{2}}{ \left( 4\,x + \frac{1}{5}\,x^3 \right) \,\mathrm{d}x } \\ &= 2\,\pi\,\left[ 2\,x^2 + \frac{1}{20}\,x^4 \right]_1^{\frac{3}{2}} \\ &= 2\,\pi\,\left\{ \left[ 2\,\left( \frac{3}{2} \right) ^2 + \frac{1}{20}\,\left( \frac{3}{2} \right) ^4 \right] - \left[ 2\,\left( 1 \right) ^2 + \frac{1}{20}\,\left( 1 \right) ^4 \right] \right\} \\ &= 2\,\pi\, \left\{ \left[ 2\,\left( \frac{9}{4} \right) + \frac{1}{20}\,\left( \frac{81}{16} \right) \right] - \left( 2 + \frac{1}{20} \right) \right\} \\ &= 2\,\pi \, \left( \frac{9}{2} + \frac{81}{320} - \frac{41}{20} \right) \\ &= 2\,\pi\,\left( \frac{1440}{320} + \frac{81}{320} - \frac{652}{320} \right) \\ &= 2\,\pi\,\left( \frac{869}{320} \right) \\ &= \frac{869\,\pi}{160} \,\textrm{units}^3 \end{align*}