volume of the solid generated by $y^{2}+z^{2}=1 ,y= \frac{1}{2}$ and y greater than equal to $0$ about $z$ axis

Question

volume of the solid generated by $y^{2}+z^{2}=1$ ,$y= \frac{1}{2}$ and $y$ greater than equal to 0 about $ z axis $

I have tried by disk method but able to get answer . I have got this figure

i have tried breaking region from $z=\frac{\sqrt{3}}{2}$

$\int_{0}^{\frac{\sqrt{3}}{2}}\pi (\frac{1}{2})^{2}dz$+$\int_{\frac{\sqrt{3}}{2}}^{1}\pi(\sqrt{1-z^{2}})^{2}dz$

$=\frac{\sqrt{3}\pi}{8}+\frac{2}{3}+\frac{3\sqrt{3}}{8}$

$=\frac{\sqrt{3}\pi}{4}+\frac{2}{3}$

Please help .(this question was asked in IIT JAM2018)

Answer 1

Note that the strip $AB$ should be on the right of the vertical line $y = \frac{1}{2}$

Consider the the elementary strip of width $dy$, whose end points are $A (y, \sqrt{1-y^2})$ and $B(y, -\sqrt{1-y^2})$

When $AB$ is revolved about the $z$ axis, the volume of the solid generated is $2 \pi y (2\sqrt{1-y^2}) dy = (-2 \pi) \sqrt{1-y^2} (-2y) dy$

Hence the volume of the solid is $$\int_{\frac{1}{2}}^1(-2 \pi) \sqrt{1-y^2} (-2y) dy = \left[\frac{4 \pi}{3}\left(1-y^2\right)^{\frac{3}{2}}\right]_1^{\frac{1}{2}} = \frac{\pi \sqrt{3}}{2}$$

Answer 2

Your integral calculation is wrong. Let's start with that. First of all, you lost some $\pi$ factors. Also, why is the lower limit of the integral $0$? You can go to negative $z$. Due to symmetry, the answer should be exactly twice as if you integrate from $0$. If you solve correctly your integral, you get $$\frac V2=\frac{2\pi}3-\frac{\pi\sqrt 3}{4}$$ or $$V=\frac{4\pi}3-\frac{\pi\sqrt 3}{2}$$ I've obtained the same result using cylinders of radius $y$, thickness $dy$ and height $2\sqrt{1-y^2}$. So the results are consistent.

Now going back to the formula, the value from answer (C) is $\frac{4\pi}3-V$. It means that it's the complementary volume. And if you read the problem again, you might realize that the area that you rotate is not given by $y<1/2$ but by $y>1/2$. This step is not intuitive. Just bad wording. So to set up correctly the integration you need to use rings of thickness $dz$, with inner radius $1/2$ and outer radius $\sqrt{1-z^2}$ $$V=\int_{-\sqrt 3/2}^{\sqrt3/2}\pi(1-z^2-\frac1{2^2})dz$$

Answer 3

First, a few things that are wrong in your calculation.

• In "$\frac{\sqrt{3}\pi}{8}+\color{red}{\frac{2}{3}+\frac{3\sqrt{3}}{8}}$", the last two terms, which come from evaluating the second integral, are missing a coefficient of "$\pi$" from the integral.

• Then you somehow combined the two terms "$\frac{\sqrt{3}\pi}{8}+\cdots+\frac{3\sqrt{3}}{8}$" into "$\color{red}{\frac{\sqrt{3}\pi}{4}}$", which is a complete mystery to me. In your previous line they aren't even like terms, since one has a $\pi$ in it and the other one doesn't. Of course, if you correct the earlier mistake, then they are supposed to be like terms, but they would add up to something different.

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Now, the actual issue is that the wording of the problem is (in my opinion) terribly ambiguous. They should've phrased it much more clearly! It appears to me that they mean the region inside the circle that lies to the right of the line $y=\dfrac{1}{2}$, not to its left. See here:

Therefore, applying the disks/washers method here will have to be in the form of the washers rather than disks because of the hole in the middle. The integration will go from $z=-\dfrac{\sqrt{3}}{2}$ to $z=\dfrac{\sqrt{3}}{2}$, like this: $$\int_{-\sqrt{3}/2}^{\sqrt{3}/2}\pi\left(\sqrt{1-z^2}\right)^2\,dz-\int_{-\sqrt{3}/2}^{\sqrt{3}/2}[\text{the integral for the hole}].$$