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Competency in Focus: Application of Calculus (Volume of Revolution)

This is problem from IIT JAM 2018 is based on calculation of volume of revolution.

Next understand the problem

Consider the region $D$ on $yz$ plane and bounded by the line $y=\frac{1}{2}$ and the curve $y^{2}+z^{2}=1$ where $y\geq0$. If the region $D$ is revolved about the $z-$axis then the volume of resulting solid is $(A)\frac{\pi}{\sqrt{3}}\qquad(B)\frac{2\pi}{\sqrt{3}}$ $\qquad(C)\frac{\pi\sqrt{3}}{2}\qquad(D)\pi\sqrt{3}$
Source of the problem
IIT JAM 2018, Question number 19
Key Competency
Calculation of volume of surface revolution
7/10
Suggested Book

Do you really need a hint? Try it first!
Imagine that we have a portion of a curve. $y=f(x)$ from $x=a$ to $x=b$. In the $xy-plane$ we revolve it around a straingh line –  $x$-axis.  The result is called solid of revolution Here in our next hint we will find techniques to calculate the volumes of solid of revolution.
Let’s construct a narrow rectangle of base width $\mathrm d x$ and height $f(x)$ sitting under the curve. Whent this rectangle is revolved around the $x-\text{axis}$. We get a disk whose radius is $f(x)$ and height is $\mathrm d x$. The volume of this disk $\mathrm d V=\pi[f(x)]^{2}\mathrm d x$. So the total volume of the solid is  $V=\int_a^b \mathrm d V=\int_a^b \pi[f(x)]^{2}\mathrm d x$. If the curve revolved around the vertical line (such as $y$-axis), then horizontal disks are used. If the curve can be solved for $x$ in terms of $y$. $x=g(y)$ then the formula would be. $V=\int_a^b[g(y)]^2\mathrm d y$. Now can you guess if the region between two curves is revolved around the axis.
Sometimes the region between two curves is revolved around the axis and a gap is created between the solid and the axis. A rectangle within the rotated region will become a disk with a hole in it, also known as washer. If the rectangle is vertical and extends from the curve $y=g(x)$ up to the curve $y=f(x)$, then when it is rotated around $x$- axis, it will result in a washer with volume equal to $\mathrm d V=\pi\{[f(x)]^{2}-[g(x)]^{2}\}\mathrm d x$. Which gives us  $V=\int_a^b \pi \{[f(x)]^{2}-[g(x)]^{2}\}\mathrm d x$.
Now in the given problem replace $x$ by $z$ , in the obove discussion $y=f(z)=\sqrt{1-z^{2}}, y=g(z)=\frac{1}{2}$ and the line $y=\frac{1}{2}$ intersects the curve at $(\frac{-\sqrt{3}}{2},\frac{1}{2})$ , $(\frac{\sqrt{3}}{2},\frac{1}{2})$ in terms of $(z,x)$ co-ordinate. and hence the volume is  $V=\pi \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}[(1-z^{2})-\frac{1}{4})]\mathrm d z$. $=\pi[\frac{3z}{4}-\frac{z^{3}}{3}]_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}$ $=\pi(\frac{3\sqrt{3}}{8}-\frac{3\sqrt{3}}{24})\times 2$ $=\pi \frac{3\sqrt{3} \times 2 \times 2}{24}$ $=\frac{\pi \sqrt{3}}{2}$

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