# Find the volume of solid generated by rotating this sector?

Okay so the question goes:

Find the volume of the solid generated by revolving the plane region bounded from upward by circle $x^2 + y^2 = 4$ and from downward by two straight lines $y = x$ and $y = -x$ around $x-axis$

So as I read this the question, I imagined he's asking me to draw a circle whose radius is 2, and then 2 lines making a sector in that circle and then revolve that sector around x-axis. I had 3 attempts at this and all 3 gave different answers.

First attempt, the original Calculus attempt (the problem is in a Calculus textbook): $$\pi \left(\int _{-2}^0\left(\sqrt{4-x^2}+x\right)^2dx\:+\:\int _0^2\left(\sqrt{4-x^2}-x\right)^2dx\:\:\right) = \frac{16\pi}{3}$$ I suspected this answer because I tried rotating the whole circle and found that the volume of the sphere was $\frac{32\pi}{3}$ which made me confused because I don't know if the rotated sector can have half the volume of the rotated sphere.

Second attempt, tried using some basic geometry:

We know that the two lines forming the sector are $y = x$ and $y = -x$, so it's pretty obvious that the angle between each one and the $x-axis$ is $45$ degrees, therefore the angle between the two lines is $90$ degrees. Which is again obvious since one is only the negative of the other. I googled the formula of the Area of Sector and it was $(\frac{\theta}{2})*r^2$, andI know the volume of any regular shape is $V = Area*h$ I didn't know whether to use the radius as the height or the height of sector so I tried both, and they evaluated to $V = 2\pi$ and $V \approx 1.8$

I suspected this one was wrong because the values are too small?

Third attempt, I simply calculated the total volume of sphere and divided by 4 knowing the sector is a quarter-circle:

$V = \frac{4}{3} \pi r^3$, $V = \frac{4}{3} \pi *(2)^3 = \frac{32\pi}{3}, V of sector = \frac{\frac{32\pi}{3}}{4} = \frac{8\pi}{3}$

this caught my attention because it was exactly the half of the first answer. But honestly, after these attempts all with different answers I just think I am completely wrong with the three of them at this point. I hope someone clears up why I am wrong, if I am.

Referring to the diagram below the exercise is to find the volume of the solid of revolution when revolving the blue region about the $x$-axis.

If one varies $x$ between the values of $-\sqrt{2}$ and $\sqrt{2}$ and uses the annulus method then

\begin{eqnarray} V&=&\int_{-\sqrt{2}}^{\sqrt{2}}\pi(R^2-r^2)\,dx\\ &=&2\pi\int_0^{\sqrt{2}}R^2-r^2\,dx\tag{1} \end{eqnarray}

where $R=\sqrt{4-x^2}$ and $r=x$. This gives a value

$$V=\frac{16\sqrt{2}\pi}{3}$$

As a check we can also find the volume of the solid of revolution formed by revolving the red region about the $x$-axis. The blue plus red volumes should sum to the volume $V=\dfrac{32\pi}{3}$.

The "red volume" can be found by the cylindrical shell method. We will use the sector on the right and double the result so that

$$V=2\int_0^\sqrt{2}2\pi rh\,dy$$

where $r=y$ and $h=\sqrt{4-y^2}-y$. So

$$V=4\pi\int_0^\sqrt{2}y\sqrt{4-y^2}-y^2\,dy$$

which has a value of

$$V=\frac{32\pi}{3}-\frac{16\sqrt{2}\pi}{3}$$

And we see that the "blue volume" and "red volume" sum correctly to the volume of the sphere.

Addendum: Note that if one integrates equation $(1)$ over the range $[-2,2]$ that will give the volume of the solid of revolution formed by revolving both the green and the blue regions (with the green volume subtracted from the blue volume), which is not the volume asked for in the question.

• Why the $sqrt(2)$? I thought the limits of the integral would only be $-2,2$ since the radius is $sqrt(4) = 2$? I plugged $2$ instead of $sqrt(2)$ in your solution and it gave my first answer, $\frac{16\pi}{3}$. – Eyad H. Feb 27 '17 at 19:06
• The limits of the integral over the blue region extend from the smallest to the largest $x$ coordinate of points in the blue region. This occurs at the points where the graphs $y=-x$ and $y=x$ intersect the circle. The $x$-coordinates of those points are $\pm\sqrt{2}$. – John Wayland Bales Feb 27 '17 at 19:11
• I have modified the graph and added additional information to explain why it is incorrect to integrate from $-2$ to $+2$. I hope this helps you to understand the issue. – John Wayland Bales Feb 27 '17 at 19:29