Set up a definite integral for the volume obtained by rotating the region between the curve $y^{2}=x$ and $y^{2}=2(x-1)$ about $y=3$

Question

So I have a similar question from a bit before:

Set up a definite integral that represents the volume obtained by rotating the region between the curve $y^{2}=x$ and $y^{2}=2(x-1)$ about the line $y=3$

I get $\int\limits\limits_{-\sqrt{2}}^{\sqrt{2}} \pi((\frac{ (3-y)^2+2 }{ 2 })^{2})-\ (3-y)^4)dy$

Is that right?

Answer 1

Using the method of washers, the washers would be perpendicular to the axis of rotation, thus the integration would be performed along the $x$-axis. However, because a line perpendicular to the axis of rotation (i.e. parallel to the $y$-axis) can intersect the region of interest in two disjoint intervals, we see that such an approach is not optimal.

Instead, the method of cylindrical shells is better suited. In such a case, the integration would be performed along the $y$-axis, since the shells are nested within each other such that they all share the line $y = 3$ as the common axis. We also note that for a representative shell, the "height" is measured as the horizontal distance between the two parabolic arcs and that horizontal lines never intersect the region of interest along more than one contiguous interval.

To this end, we first solve for the $y$-coordinates for which the parabolas intersect: $$y^2 = x, \quad y^2 = 2(x-1)$$ implies that $(x,y) = (2, \pm \sqrt{2})$. Then for a representative shell of radius $r(y) = 3-y$, the height of such a shell is $$h(y) = (y^2/2 + 1) - y^2 = 1 - y^2/2,$$ and the differential volume is $$dV = 2\pi r(y) h(y) \, dy = 2\pi (3-y)(1 - y^2/2) \, dy.$$ The total volume is then $$V = \int_{y=-\sqrt{2}}^{\sqrt{2}} \, dV = 2\pi \int_{y=-\sqrt{2}}^{\sqrt{2}} (3-y)(1-y^2/2) \, dy.$$

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If we are interested in the volume of the same region if rotated about the $y$-axis (i.e. $x = 0$), then the method of washers would apply, but now the outer radius corresponds to the curve $y^2 = 2(x-1)$ and the inner corresponds to the curve $y^2 = x$, even though the cross-sectional intervals are the same as in the previous computation. Thus the differential volume of a representative washer is $$dV = \pi((y^2/2 + 1)^2 - y^4) \, dy,$$ and the volume is $$V = \int_{y=-\sqrt{2}}^{\sqrt{2}} dV = \pi \int_{y=-\sqrt{2}}^{\sqrt{2}} -\frac{3y^4}{4} + y^2 + 1 \, dy.$$

Answer 2

OK, you must really sketch your region before deciding which method to use. Sometimes it doesn't matter and sometimes one is much easier than the other.

In this case, without changing the coordinate plane (you could simply switch the $x$ and $y$ coordinate in this problem, then revolve around $x = 3$ and this this would be a very easy problem to sketch in what you normally think of as the $xy$-plane).

It helps to realize that the $y^2 = x$ is a "sideways" parabola. It helps even further to realize that $y^2 = 2(x -1)$ is also a sidways parabola but shifted and stretched/compressed:

The sketch will show that the shell method is most appropriate here. The washer method would work but it would require breaking the region into (at least) two parts.

Once the region is sketched, it's a matter of finding $r$, $h$, and $dr$. $dr$ is easy, it's clearly $dy$. $r$ is also fairly simple, it's clearly $3 - y$--which is nice since we have $dy$, this is already in the form we need. Finally $h = \Delta x = x_R-x_L$. $x_R$ is given by $y^2 = 2(x-1)$ and $x_L$ is given by $y^2 = x$ therefore:

$$ x_L = y^2 $$

and

$$ y^2 = 2(x_R - 1) \\ x_R = \frac{y^2}{2} + 1 $$

Therefore:

\begin{align} \Delta x =&\ x_R - x_L\\ =&\ \frac{y^2}{2} + 1 - y^2 \\ =&\ -\frac{y^2}{2} + 1 \end{align}

Summarizing this gives:

$$ r = 3-y \\ h = -\frac{y^2}{2} + 1 \\ dr = dy \\ dV = 2\pi rh dr = 2\pi (3-y)\left(1-\frac{y^2}{2}\right)dy $$

Finally we must find the limits of integration what $y$ values does this region range from? Simple, it's where $\Delta x = 0$:

$$ 1 - \frac{y^2}{2} = 0 \\ y^2 = 2 \rightarrow y = \pm \sqrt{2} $$

So finally the integral should be:

$$ 2\pi\int\limits_{-\sqrt{2}}^{\sqrt{2}}\left((3-y)\left(1-\frac{y^2}{2}\right)\right)dy $$

Answer 3

Rotating about the $y=3$ $$V=\pi\int^{2}_{0} ((3-\sqrt x)^2-(3-\sqrt{2x-1})^2)dx$$

Answer 4

I wrote the following just to offer an alternative. Yet please do observe that at the end I get an extra $\;-2\pi\;$ instead of the $\;8\sqrt2\pi\;$ that you get doing the calculations as in the other answers. I'd like to know where I did go wrong:

Since the plane region that is going to be revolved about a line parallel to the $\;x\,-$ axis is symmetric with respect to this axis, I'd rather have the region revolved about the line $\;y=-3\;$ .

Thus, I can revolve the region $\;0\le x\le1\,- $ axis, with the functions:

$$\begin{cases}\text{Above:}\;\;y_1=\sqrt x+3\\{}\\\text{Below:}\;\;y_2=-\sqrt x+3\end{cases}\;\;,\;\;0\le x\le1\implies V_0=\pi\int_0^1(y_1^2-y_2^2)dx\implies$$

$$V_0=\pi\int_0^112\sqrt x\;dx=12\pi\cdot\frac23\cdot1^{3/2}=8\pi$$

and now the region $\;1\le x\le2\;$ , with the sum of the volumes for the functions:

$$V_1:\;y\ge 0\implies\;\begin{cases}\text{Above:}\;\;y_1=\sqrt x+3\\{}\\\text{Below:}\;\;y_2=\sqrt{2x-2}+3\end{cases}\;\;,\;\;1\le x\le2\implies V_1=\pi\int_0^1(y_1^2-y_2^2)dx$$

$$\implies V_1=\pi\int_1^2\left[-x+6\left(\sqrt x-\sqrt{2x-2}\right)\right]dx=$$

$$=\pi\left(\left.-\frac12x^2\right|_1^2+6\left(\left.\frac23x^{3/2}\right|_1^2-\left.\frac12\frac23(2x-2)^{3/2}\right|_1^2\right)\right)=$$

$$=\pi\left(-\frac32+6\left(\frac23(2\sqrt2-1)-\frac13(2\sqrt2-0)\right)\right)=\pi\left(-\frac32+4\sqrt2-4\right)=\left(4\sqrt2-\frac{11}2\right)\pi$$

and

$$V_2:\;y\le 0\implies\;\begin{cases}\text{Above:}\;\;y_1=-\sqrt{2x-2}+3\\{}\\\text{Below:}\;\;y_2=-\sqrt x+3\end{cases}\;,\;1\le x\le2\implies V_2=\pi\int_0^1(y_1^2-y_2^2)dx$$

$$\implies V_2=\pi\int_1^2\left[x-2+6\left(\sqrt x-\sqrt{2x-2}\right)\right]dx=$$

$$\pi\left[\left.\frac12x^2\right|_1^2-2+6\left(\left.\frac23x^{3/2}\right|_1^2-\left.\frac13(2x-2)^{3/2}\right|_1^2\right)\right]=$$

$$=\pi\left(\frac32-2+6\left(\frac23(2\sqrt2-1)-\frac13(2\sqrt2-0)\right)\right)=$$

$$=\pi\left(-\frac12+4\sqrt2-4\right)=\left(4\sqrt2-\frac92\right)\pi$$

so the final volume is

$$V=V_0+V_1+V_2=\left(8\left(1+\sqrt2\right)-10\right)\pi=\left(8\sqrt2-2\right)\pi$$

Answer 5

Using the shell method,

$\hspace{.2 in}\displaystyle V=\int_{-\sqrt{2}}^{\sqrt{2}}2\pi r(y) h(y)dy=\color{blue}{\int_{-\sqrt{2}}^{\sqrt{2}}2\pi (3-y)\left(\left(\frac{y^2}{2}+1\right)-y^2\right)dy}$

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$\;\;\;$Alternate answer: (disc method)

$\hspace{.2 in}\displaystyle V=\color{green}{\int_0^{2}\pi\left((3+\sqrt{x})^2-(3-\sqrt{x})^2\right)dx-\int_1^{2}\pi\left(\left(3+\sqrt{2(x-1)}\right)^2-\left(3-\sqrt{2(x-1)}\right)^2\right)dx}$