A region A is bounded by y = 1 x , y = 0 , x = 1 , and x = 4 . A is revolved around...

Question:

A region {eq}A {/eq} is bounded by {eq}y = \frac{1}{x}, \ y = 0, \ x = 1, {/eq} and {eq}x = 4 {/eq}. {eq}A {/eq} is revolved around the {eq}y {/eq}-axis. Set up the integral representing the volume of the solid. Find the volume of the solid.

We will be using the Disk Method.

Let us consider a cross-section whose radius varies and it is equal to the y coordinate on the given curve.

{eq}\text{ }\\ \displaystyle r=y= \frac{1}{x} \\ \text{If the width of this disk is }\\ dx\\ \text{then its volume will be }\\ \displaystyle dV=\pi r^2 dx\\ \displaystyle \Rightarrow dV=\pi \left( \frac{1}{x} \right)^2 dx\\ \displaystyle \Rightarrow dV=\pi \left( \frac{1}{x^2} \right) dx\\ {/eq}

The limit of integration is:

{eq}\displaystyle 1\le x \le 4\\ {/eq}

Total volume of the solid is represented by the following integral:

{eq}\displaystyle V=\pi \int_1^4 \left( \frac{1}{x^2} \right) \, dx\\ {/eq}

Now integrate this differential volume to get the total volume of revolution.

{eq}\begin{align} V&=\pi \int_1^4 \left( \frac{1}{x^2} \right) \, dx\\ \Rightarrow V&=\pi \left[-\frac{1}{ x} \right]_1^4 \\ &=\pi \left[-\frac{1}{ 4} \right]-\pi \left[-\frac{1}{ 1} \right]\\ &=\pi-\frac{\pi }{ 4}\\ &=\frac{3\pi }{ 4}\\ \end{align} {/eq}

Therefore, the volume of the solid is: {eq}\displaystyle \boxed{\mathbf{V=\frac{3\pi}{ 4}}} {/eq}