Brenton's questions via email about volume by revolution

In summary, volume by revolution is a mathematical concept used to find the volume of a three-dimensional shape created by rotating a two-dimensional shape around an axis. This is done using the formula V = π ∫<sub>a</sub><sup>b</sup> (f(x))<sup>2</sup>dx. It has various real-life applications, but can only be applied to symmetrical shapes and assumes a smooth surface. To solve real-world problems, one must identify the axis of rotation and the two-dimensional shape and use the correct bounds for the integral. Practice and understanding the problem are key in successfully using this concept.
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Q5. Here is a graph of the region to be integrated and the line to be rotated around.

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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.

View attachment 5642

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

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

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In conclusion, the volumes of the solid in Q5 and Q6 are $\displaystyle \begin{align*} \frac{171\,\pi}{32}\,\textrm{units}^3 \end{align*}$ and $\displaystyle \begin{align*} \frac{869\,\pi}{160} \,\textrm{units}^3 \end{align*}$, respectively. These volumes can be calculated by using the formula for volume of a solid of revolution, which is $\displaystyle \begin{align*} V = \pi\,\int_a^b{f(x)^2\,\mathrm{d}x} \end{align*}$, where $\displaystyle \begin{align*} f(x) \end{align*}$ is the function that defines the shape of the solid, and $\displaystyle \begin{align*} a \end{align*}$ and $\displaystyle \begin{align*} b \end{align*}$ are the limits of integration. It is important to accurately visualize the solid and choose the correct axis of rotation before setting up the integral.
 

FAQ: Brenton's questions via email about volume by revolution

1. What is volume by revolution?

Volume by revolution is a mathematical concept that involves finding the volume of a three-dimensional shape created by rotating a two-dimensional shape around an axis. This is commonly used in calculus and physics to solve problems involving rotational motion and fluid dynamics.

2. How do you calculate volume by revolution?

The formula for calculating volume by revolution is V = π ∫ab (f(x))2dx, where V is the volume, π is the mathematical constant pi, and f(x) is the function that represents the two-dimensional shape being rotated. The integral is taken over the interval [a,b].

3. What are some real-life applications of volume by revolution?

Volume by revolution has many practical applications, such as finding the volume of a soda can or a cylindrical container, calculating the amount of oil in a pipeline, and determining the volume of a liquid in a container that is being stirred or rotated.

4. Are there any limitations to using volume by revolution?

One limitation of using volume by revolution is that it can only be applied to symmetrical shapes. Additionally, the formula assumes that the shape being rotated is continuous and has a smooth surface. If the shape is irregular or has discontinuities, the calculated volume may not be accurate.

5. How can I use volume by revolution to solve real-world problems?

To use volume by revolution to solve a real-world problem, you first need to identify the axis of rotation and the two-dimensional shape that will be rotated. Then, you can use the formula V = π ∫ab (f(x))2dx to calculate the volume. It is important to understand the problem and choose the correct bounds of the integral to get an accurate result. Practice and familiarizing yourself with different types of problems can also help improve your problem-solving skills.

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