# Kishan's question via email about volume by revolution

• MHB
• Prove It
In summary, the volume obtained by rotating the region \displaystyle \begin{align*} R \end{align*}, bounded above by \displaystyle \begin{align*} y = 2 - 5\,\sqrt{x} \end{align*} and below by \displaystyle \begin{align*} y = 0 \end{align*} in the first quadrant, around the line \displaystyle \begin{align*} y = -4 \end{align*} is \displaystyle \begin{align*} \frac{24\,\pi}{25}\,\textrm{units}^3 \end{align*}.
Prove It
Gold Member
MHB
\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 realize 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*}.

chwala

## 1. What is "volume by revolution"?

"Volume by revolution" refers to the process of finding the volume of a three-dimensional object by rotating it around a specific axis. This is typically done using mathematical formulas and integrals.

## 2. How is volume by revolution calculated?

The exact formula for calculating volume by revolution depends on the shape of the object and the axis of rotation. However, it typically involves setting up an integral and integrating the area of cross-sections of the object as it is rotated around the axis.

## 3. What is the practical application of volume by revolution?

Volume by revolution is often used in engineering and design to calculate the volume of objects such as pipes, cylinders, and curved surfaces. It can also be used in physics to determine the volume of rotating objects, such as planets or stars.

## 4. Can volume by revolution be calculated for irregular shapes?

Yes, volume by revolution can be calculated for irregular shapes as long as the object has a consistent cross-sectional shape and the axis of rotation is known. In these cases, the integral may be more complex, but the same principles apply.

## 5. Are there any limitations to using volume by revolution to calculate volume?

Yes, volume by revolution assumes that the object being rotated is solid and has a consistent cross-sectional shape. If the object is hollow or has varying cross-sections, a different method of calculating volume may be more appropriate.

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