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I am to find the volume of the area $R$ bounded by the curve $x=y^2+2$, $y=x-4$ and $y=0$. I have already found the points of intersection by first setting the lines equal to each other and used the quadratic formula: \begin{align*} y^2+2=y+4 \\-y^2+y+2=0 \\ \\y_{1,2}=\frac{1\pm3}{2}\\ \\y_1 = 2 \\y_2=-1 \end{align*}

With regard to y:

\begin{align*} A=\int_0^2{(y+4)-(y^2+2)}dy\\ A=\left[(2\cdot2)+\frac{2^2}{2}-\frac{2^3}{3}\right]-\left[(2\cdot0)-\frac{0^2}{2}-\frac{0^3}{3}\right]\\ A=4+2-\frac{8}{3}\\ A=\frac{10}{3} \end{align*}

Then I tried finding the volume of $R$:

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We have:

$f(y)=2+y-y^2 dy $ \begin{align*} V= 2\pi\int_a^byf(y) dy\\ =2\pi\left(\int_2^4 y^3+2ydy\right)+2\pi\left(\int_4^6(y^3+2y)-(y^2+4y)dy\right)\\ =4\pi\left(\left[\frac{y^4}{4}+y^2\right]_2^4 + \left[\left(\frac{y^4}{4}+y^2\right)-\left(\frac{y^3}{3}+2y^2\right)\right]_4^6\right)\\ =4\pi \cdot 290\\ = \frac{\pi}{2}(145) \end{align*}

But the right answer should be $16\frac{\pi}{3}$. What am I doing wrong?

Théophile
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Mampenda
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  • Why so many upvotes?! – Parcly Taxel Nov 09 '20 at 17:33
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    @ParclyTaxel In MathSE the upvotes are given when the OP in his post has shown his attempts to solve the problem and his question is clear about where he is stuck. I guess that's why he has so many upvotes. –  Nov 09 '20 at 17:36
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    This is a nicely presented question. A couple of comments: first, you calculate the area $A$ of the region to be revolved around the $x$-axis. Is that necessary? You don't use it later. Second, when calculating the volume, you have $y$ going from $2$ to $4$ in one integral, and from $4$ to $6$ in the other. You've already established, however, that $y$ gets no greater than $2$ ... – Théophile Nov 09 '20 at 17:49

1 Answers1

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Assuming $R$ is the region in the first quadrant (i.e. with $x\ge0$ and $y\ge0$), then the volume - using cylindrical shells - is

$$\begin{align} V&=2\pi\int_0^2y((y+4)-(y^2+2))\,\mathrm dy\\[1ex] &=2\pi\int_0^2(2y+y^2-y^3)\,\mathrm dy\\[1ex] &=2\pi\left[y^2+\frac{y^3}3-\frac{y^4}4\right]_0^2\\[1ex] &=2\pi\left(2^2+\frac{2^3}3-\frac{2^4}4\right)=\boxed{\frac{16\pi}3} \end{align}$$

Your error is in the limits of integration. You seem to be using the bounds for the variable $x\in[2,6]$, not $y\in[0,2]$.

If you did want to set up an integral using the bounds for $x$, you can use disks and washers:

$$\begin{align} V&=\pi\left(\int_2^4\left(\sqrt{x-2}\right)^2\,\mathrm dx+\int_4^6\left(\left(\sqrt{x-2}\right)^2-(x-4)^2\right)\,\mathrm dx\right) \end{align}$$

and you would find this to have the same value.

user170231
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