## What are we learning ?

**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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