# 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 $y=3$

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?

• It would appear you are using the washer method, is this correct? If so why are you choosing this method? (I'm not saying it's wrong, I'm only asking for more of your input). Why are you choosing this method over the shell method? – Jared Jul 3 '16 at 5:58
• I do not have the solutions so I am not sure if my answer is correct. I am using the washer method. There is no reason. Since I am rotating about the y-axis it just felt like it made more sense to me. I'm not really sure how I would have used the shell method. – Future Math person Jul 3 '16 at 5:59
• When using the washer method you must take a cut through the region perpendicular to the axis of rotation. In this case that would mean you would integrate with respect to $x$. If you wish to integrate wrt $y$ you would use the cylindrical shell method $\int 2\pi rh\,dy$ with $r$ and $h$ functions of $y$. – John Wayland Bales Jul 3 '16 at 6:01
• Well, I am guessing then, the radius would just we what I have inside the Integrand (But not squared of course), and the height would just be y? – Future Math person Jul 3 '16 at 6:07
• No, your integrand is completely wrong, even without the squares. – John Wayland Bales Jul 3 '16 at 6:11

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 $$V = \int_{y=-\sqrt{2}}^{\sqrt{2}} \, dV = 2\pi \int_{y=-\sqrt{2}}^{\sqrt{2}} (3-y)(1-y^2/2) \, dy.$$

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 $$V = \int_{y=-\sqrt{2}}^{\sqrt{2}} dV = \pi \int_{y=-\sqrt{2}}^{\sqrt{2}} -\frac{3y^4}{4} + y^2 + 1 \, dy.$$

• Well, your animations kick my answer's ass, lol. – Jared Jul 3 '16 at 6:46
• Both of your explanations were really helpful! This is embarrassing for me because I took Calc II years ago and now I''m taking a class on PDE's xD . I always hated volumes and surfaces of revolution. Thanks so much for the detailed explanation both of you! If I JUST rotated about the y-axis instead of y=3, r would just be y and not 3-y I assume? – Future Math person Jul 3 '16 at 6:50
• @SubhashisChakraborty If you rotate about the $y$-axis, then you'd be doing washers and not shells, because the line $y = 3$ is parallel to the $x$-axis. So perhaps you intended to ask, "if I just rotated about the $x$-axis instead of $y = 3$,...." In which case, the answer to that question is yes. This is because the radius function expresses the radius of a representative shell as a function of the $y$-coordinate. If the axis of rotation was $y = 0$ (the $x$-axis), then $r(y) = y$. – heropup Jul 3 '16 at 6:53
• Could I just not use shells and say that the volume is given by: $V = \int_{y=-\sqrt{2}}^{\sqrt{2}} \, dV = 2\pi \int_{y=-\sqrt{2}}^{\sqrt{2}} (y)(1-y^2/2) \, dy$ ? Sorry for asking this. I am really bad at visualizing this so I have to >.< . – Future Math person Jul 3 '16 at 6:58
• @SubhashisChakraborty I think you mean the $x$-axis (not the $y$-axis). If so, then still the shell method would be preferable because, if you used the washer method, there would be a region which went from $y = 0$ to $y = \sqrt{x}$ and then a separate region which went from $y = \sqrt{2(x-1)}$ to $y = \sqrt{x}$ as opposed to simply giving the same integral as $\int\limits_0^{\sqrt{2}}\left(y\left(1-\frac{y^2}{2}\right)\right)dy$...the limits are slightly different because of the symmetry of the problem (when rotated about the $x$-axis). – Jared Jul 3 '16 at 7:14

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:

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

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

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

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

Using the shell method,

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

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

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