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I'm asking this question for a friend who is working on a volumes of revolution question (she is not familiar with stackexchange, and I'm not particularly good at volumes of revolution problems). The question is given as:

Calculate the volume of revolution of $R$ where $R$ is the region bounded by $y=5x-x^{2}$, $y=3+x$ rotated about $a)$ $y=3$, $b)$ $x=3$

Apart from having trouble solving the question, she has trouble visualizing what this question would look like on a graph. Any help would be appreciated!

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  • $\begingroup$ Drawing a graph of the line and the quadratic, paying attention to the points where they intersect, will bring you closer to visualizing the region $R$ $\endgroup$ – John Joy Jun 5 '16 at 4:06
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First off, we make a sketch of the region $R$.

Figure 1

We can find the intersection of the two curves by setting $$y=5x-x^2=3+x$$ So that $x^2-4x+3=(x-1)(x-3)=0$ and the points of intersection are $(1,4)$ and $(3,6)$. To find the volume of the solid of revolution about $y=3$, it will be easiest to integrate washers: $$\begin{align}V&=\int\pi\left(r_{\text{outer}}^2-r_{\text{inner}}^2\right)dh=\int_1^3\pi\left[(5x-x^2-3)^2-(3+x-3)^2\right]dx\\ &=\int_1^3\pi\left[x^4-10x^3+30x^2-30x+9\right]dx=\pi\left[\frac{x^5}5-\frac{5x^4}2+10x^3-15x^2+9x\right]_1^3\\ &=\pi\left[\frac{242}5-200+260-120+18\right]=\frac{32\pi}5\end{align}$$ To find the volume of the solid of revolution about $x=3$, it will be easiest to integrate oatmeal boxes: $$\begin{align}V&=\int2\pi rh\,dr=\int_1^32\pi(3-x)\left[5x-x^2-3-x\right]dx\\ &=\int_1^32\pi\left[x^3-7x^2+15x-9\right]dx=2\pi\left[\frac{x^4}4-\frac{7x^3}3+\frac{15x^2}2-9x\right]_1^3\\ &=2\pi\left[20-\frac{182}3+60-18\right]=\frac{8\pi}3\end{align}$$

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