Problem 61 - Essential Calculus

Question

Consider the region satisfying the inequalities. (a) Find the area of the region. (b) Find the volume of the solid generated by revolving the region about the \(x\) -axis. (c) Find the volume of the solid generated by revolving the region about the \(y\) -axis. $$ y \leq e^{-x}, y \geq 0, x \geq 0 $$

Step-by-Step Solution

Answer

The area of the region is 1. The volume of the solid generated by revolving the region about the x-axis is \(\frac{\pi}{2}\) and the volume of the solid when revolved around the y-axis is \(\pi\)

Step 1: Find the Area

Set up the integral to represent the area under the curve from \(x=0\) to \(x=+\infty\). We can find the area using the integral: \(Area = \int_{0}^{+\infty} e^{-x} dx\). This is an improper integral, and we use limit to solve it.

Step 2: Evaluate the Area Integral

To evaluate this integral, we can understand it as a limit: \(Area = \lim_{b \to +\infty} \int_{0}^{b} e^{-x} dx\). Now, the antiderivative of \(e^{-x}\) is \(-e^{-x}\). So, the integral can be written as \(\lim_{b \to +\infty} [ -e^{-x} ]_0^{b}\), which evaluates to 1.

Step 3: Setup Integral for Volume when Revolved Around the x-axis

We can use the Disc Method to find the volume. \(Volume = \pi \int_{0}^{+\infty} (e^{-x})^2 dx\). Again, this is an improper integral because of the upper limit.

Step 4: Evaluate the Volume Integral

Following the same approach as in Step 2, we solve the integral and get \(Volume = \lim_{b \to +\infty} \pi \int_{0}^{b} e^{-2x} dx\). This gives us a volume of \(V = \frac{\pi}{2}\).

Step 5: Setup Integral for Volume when Revolved Around the y-axis

We use the cylindrical shells method here. \(Volume = 2\pi \int_{0}^{1} y(-\ln y) dy\). This is a definite integral since we have a bounded region.

Step 6: Evaluate the Volume Integral

Evaluate the integral with respect to \(y\). It can be solved by applying integration by parts and finding the limits. With careful calculation, we find the volume is \(V = \pi\).