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Edin's question via email about volume by revolution.

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
R is the region bounded above by $\displaystyle \begin{align*} y = 1 + x \end{align*}$ and below by $\displaystyle \begin{align*} y = 2\,x^2 \end{align*}$.

(a) If R is rotated about the line $\displaystyle \begin{align*} y = 3 \end{align*}$, construct a definite integral for the volume of the region generated.

(b) If R is rotated about the line $\displaystyle \begin{align*} x = -1 \end{align*}$ construct a definite integral for the volume of the region generated.
We should note that the two functions intersect at $\displaystyle \begin{align*} x = -\frac{1}{2} \end{align*}$ and $\displaystyle \begin{align*} x = 1 \end{align*}$.

(a) Using the method of washers, the inner radius is $\displaystyle \begin{align*} 3 - \left( x + 1 \right) = 2 - x \end{align*}$ and the outer radius is $\displaystyle \begin{align*} 3 - 2\,x^2 \end{align*}$, so the volume is

$\displaystyle \begin{align*} V &= \int_{-\frac{1}{2}}^1{ \pi \, \left( 3 - 2\,x^2 \right) ^2\,\mathrm{d}x } - \int_{-\frac{1}{2}}^1{ \pi \, \left( 2 - x \right) ^2 \,\mathrm{d}x } \\ &= \pi \int_{-\frac{1}{2}}^1{ \left[ \left( 3 - 2\,x^2 \right) ^2 - \left( 2 - x \right) ^2 \right] \,\mathrm{d}x } \\ &= \pi \int_{-\frac{1}{2}}^1{ \left[ 9 - 12\,x^2 + 4\,x^4 - \left( 4 - 4\,x + x^2 \right) \right]\,\mathrm{d}x } \\ &= \pi \int_{-\frac{1}{2}}^1{ \left( 4\,x^4 - 13\,x^2 + 4\,x + 5 \right) \,\mathrm{d}x } \end{align*}$


(b) Using the method of cylindrical shells, the radius of each cylinder is $\displaystyle \begin{align*} x + 1 \end{align*}$ and the height of each cylinder is $\displaystyle \begin{align*} x + 1 - 2\,x^2 \end{align*}$. The area of each rectangular cylindrical shell is $\displaystyle \begin{align*} 2\,\pi\,r\,h = 2\,\pi\,\left( x + 1 \right) \left( x + 1 - 2\,x^2 \right) \end{align*}$, and they will all be summed up between $\displaystyle \begin{align*} x = -\frac{1}{2} \end{align*}$ and $\displaystyle \begin{align*} x = 1 \end{align*}$, thus the volume is

$\displaystyle \begin{align*} V &= \int_{-\frac{1}{2}}^1{ 2\,\pi\,\left( x + 1 \right) \left( x + 1 - 2\,x^2 \right) \,\mathrm{d}x } \\ &= 2\,\pi \int_{-\frac{1}{2}}^1{ \left( x^2 + x - 2\,x^3 + x + 1 - 2\,x^2 \right)\,\mathrm{d}x } \\ &= 2\,\pi \int_{-\frac{1}{2}}^1{ \left( 2\,x + 1 - x^2 - 2\,x^3 \right) \,\mathrm{d}x } \end{align*}$