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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.

First look at the knowledge graph.

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
Difficulty Level
7/10
Suggested Book
Integral Calculus by Gorakh Prasad

Start with hints 

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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