- #1
Prove It
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What is the volume of the solid formed by the region bounded between the functions $\displaystyle \begin{align*} y 2\,x^2 \end{align*}$ and $\displaystyle \begin{align*} y = x + 1 \end{align*}$ being rotated around the line $\displaystyle \begin{align*} y = 3 \end{align*}$?
To start with, we should find the points of intersection of the two functions, as these will be the terminals of our regions of integration.
$\displaystyle \begin{align*} 2\,x^2 &= x + 1 \\ 2\,x^2 - x - 1 &= 0 \\ 2\,x^2 - 2\,x + x - 1 &= 0 \\ 2\,x\,\left( x - 1 \right) + 1 \,\left( x - 1 \right) &= 0 \\ \left( x - 1 \right) \left( 2\,x + 1 \right) &= 0 \\ x - 1 &= 0 \textrm{ or } 2\,x + 1 = 0 \\ x &= 1 \textrm{ or } x = -\frac{1}{2} \end{align*}$
As we are rotating around the line $\displaystyle \begin{align*} y = 3 \end{align*}$, it would make sense to move everything down by 3 units. The areas and volumes will still be exactly the same, but we will be rotating around the x axis.
The furthest function from the x-axis will be $\displaystyle \begin{align*} y = 2\,x^2 - 3 \end{align*}$, so we should evaluate the volume formed by rotating the area between that function and the x-axis between $\displaystyle \begin{align*} x = -\frac{1}{2} \end{align*}$ and x = 1 around the x axis, and then subtract the volume formed by rotating the area between the function $\displaystyle \begin{align*} y = x - 2 \end{align*}$ and the x-axis between $\displaystyle \begin{align*} x = -\frac{1}{2} \end{align*}$ and x = 1 around the x axis.
To evaluate each volume, picture each region being broken up into a number of rectangles. Each rectangle will then be rotated around the x-axis to form a cylinder. The volume of a cylinder is $\displaystyle \begin{align*} \pi\,r^2\,h \end{align*}$. Each cylinder will have a radius that is the same as the y value, and a height that is equal to $\displaystyle \begin{align*} \Delta x \end{align*}$, a small change in x.
So the outer volume we can approximate by $\displaystyle \begin{align*} \sum{ \pi\,\left( 2\,x^2 - 3 \right) ^2 \,\Delta x } \end{align*}$, and if we increase the number of cylinders (making each cylinder thinner) we get a better approximation. 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 outer volume is exactly $\displaystyle \begin{align*} \int_{-\frac{1}{2}}^1{ \pi\,\left( 2\,x^2 - 3 \right) ^2\,\mathrm{d}x } \end{align*}$. By identical arguments, the inner volume to be subtracted is exactly equal to $\displaystyle \begin{align*} \int_{-\frac{1}{2}}^1{ \pi\,\left( x - 2\right) ^2 \,\mathrm{d}x } \end{align*}$. Thus the total volume we want to evaluate is
$\displaystyle \begin{align*} v &= \int_{-\frac{1}{2}}^1{ \pi\,\left( 2\,x^2 - 3 \right) ^2\,\mathrm{d}x } - \int_{-\frac{1}{2}}^1{ \pi\,\left( x - 2 \right) ^2\,\mathrm{d}x } \\ &= \pi\int_{-\frac{1}{2}}^1{ \left[ \left( 2\,x^2 - 3 \right) ^2 - \left( x - 2 \right) ^2 \right] \,\mathrm{d}x } \\ &= \pi \int_{-\frac{1}{2}}^1{ \left[ 4\,x^4 - 12\,x^2 + 9 - \left( x^2 - 4\,x + 4 \right) \right] \,\mathrm{d}x } \\ &= \pi\int_{-\frac{1}{2}}^1{ \left( 4\,x^4 - 13\,x^2 + 4\,x + 5 \right) \,\mathrm{d}x } \\ &= \pi\,\left[ \frac{4\,x^5}{5} - \frac{13\,x^3}{3} + 2\,x^2 + 5\,x \right] _{-\frac{1}{2}}^1 \\ &= \pi\,\left\{ \left[ \frac{4\,\left( 1 \right) ^5}{5} - \frac{13\,\left( 1 \right) ^3}{3} + 2\,\left( 1 \right) ^2 + 5\,\left( 1 \right) \right] - \left[ \frac{4\,\left( -\frac{1}{2} \right) ^5}{5} - \frac{13\,\left( -\frac{1}{2} \right) ^3}{3} + 2\,\left( -\frac{1}{2} \right) ^2 + 5\,\left( -\frac{1}{2} \right) \right] \right\} \\ &= \pi \,\left\{ \left( \frac{4}{5} - \frac{13}{3} + 2 + 5 \right) - \left[ \frac{4\,\left( -\frac{1}{32} \right)}{5} - \frac{13\,\left( -\frac{1}{8} \right) }{3} + 2\,\left( \frac{1}{4} \right) - \frac{5}{2} \right] \right\} \\ &= \pi\,\left[ \left( \frac{12}{15} - \frac{65}{15} + \frac{105}{15} \right) - \left( -\frac{1}{40} + \frac{13}{24} + \frac{1}{2} - \frac{5}{2} \right) \right] \\ &= \pi\,\left[ \frac{52}{15} - \left( -\frac{3}{120} + \frac{65}{120} - \frac{240}{120} \right) \right] \\ &= \pi\,\left[ \frac{52}{15} - \left( -\frac{178}{120} \right) \right] \\ &= \pi\,\left( \frac{52}{15} + \frac{89}{60} \right) \\ &= \pi\,\left( \frac{208}{60} + \frac{89}{60} \right) \\ &= \pi\,\left( \frac{297}{60} \right) \\ &= \frac{99\,\pi}{20}\,\textrm{units}^3 \end{align*}$