Question

Find the volume of the following solids of revolution. Sketch the region in question. The region bounded by \(y=1 /\left(x^{2}+1\right), y=0, x=1,\) and \(x=4\) revolved about the \(y\) -axis

Short answer

Answer: The volume of the solid of revolution is \(V = \frac{15\pi}{2} - \frac{\pi}{2}\ln{\frac{17}{2}}\).

Step-by-step solution

Understand the problem and draw a graph

The first step is to understand the problem and visualize it by drawing a graph. The region is bounded by the curve \(y=\frac{1}{x^2+1}\), the \(y\)-axis, and the lines \(x=1\) and \(x=4\). Since we are revolving this region around the \(y\)-axis, we need to find the radius and height of each infinitesimal disk that makes up the solid.

Determine radius and height

To set up the integral, we need to determine the radius and height of each disk. Since we are revolving the region around the \(y\)-axis, the radius of each disk is the \(x\)-coordinate, and the height is given by the curve \(y=\frac{1}{x^2+1}.\) Therefore, the radius of a disk is the distance from the \(y\)-axis to a point on the curve, which is simply \(x\), and the height is given by \(y = \frac{1}{x^2+1}\).

Set up the integral for volume using the disk method

Using the disk method, the volume \(V\) of the solid can be computed as the integral of the infinitely many disks from the starting point \(x=1\) to the ending point \(x=4\). The volume of each infinitesimal disk is given by \(\pi (\text{radius})^2 (\text{height})\), which in this case, is \(\pi(x^2)\left(\frac{1}{x^2+1}\right)\). So, the integral for the volume should be: $$V = \int_{1}^{4}\pi(x^2)\left(\frac{1}{x^2+1}\right) dx$$

Evaluate the integral

Now, we will evaluate the integral: $$V = \int_{1}^{4}\pi(x^2)\left(\frac{1}{x^2+1}\right) dx = \pi\int_{1}^{4} \frac{x^2}{x^2+1} dx$$ Let's perform a substitution: \(u = x^2 + 1\), so \(\frac{du}{dx} = 2x\), which means \(dx = \frac{du}{2x}\). Now, we need to change the limits of integration. When \(x=1\), \(u=1^2+1 = 2\). When \(x=4\), \(u=4^2+1=17\). The integral is now: $$V = \pi \int_{2}^{17} \frac{x^2}{u} \frac{du}{2x}$$ Simplify the expression: $$V = \frac{\pi}{2} \int_{2}^{17} \frac{x^2}{u} du$$ Since \(x^2 = u - 1\): $$ V = \frac{\pi}{2} \int_{2}^{17} \frac{u -1}{u} du$$ We can split this into two integrals: $$ V = \frac{\pi}{2} \left( \int_{2}^{17} 1 du - \int_{2}^{17} \frac{1}{u} du \right)$$ Now, let's integrate: $$ V = \frac{\pi}{2} [ (u \big|_{2}^{17} ) - \ln{(u)}\big|_{2}^{17} ]$$ Evaluating the limits gives: $$ V = \frac{\pi}{2} [ (17 - 2) - (\ln{17} - \ln{2})]$$ Finally, simplify the result: $$ V = \frac{\pi}{2} [15 - \ln{\frac{17}{2}}]$$ So, the volume of the solid of revolution is: $$V = \frac{15\pi}{2} - \frac{\pi}{2}\ln{\frac{17}{2}}$$