Problem 64 - Calculus Early Transcendentals

Question

Let \(R\) be the region in the first quadrant bounded by the circle \(x^{2}+y^{2}=r^{2}\) and the coordinate axes. Find the volume of a hemisphere of radius \(r\) in the following ways. a. Revolve \(R\) about the \(x\) -axis and use the disk method. b. Revolve \(R\) about the \(x\) -axis and use the shell method. c. Assume the base of the hemisphere is in the \(x y\) -plane and use the general slicing method with slices perpendicular to the \(x y\) -plane and parallel to the \(x\) -axis.

Step-by-Step Solution

Answer

Question: Find the volume of a hemisphere of radius \(r\) in the first quadrant bounded by the circle \(x^2 + y^2 = r^2\) and the coordinate axes using three different methods: the disk method with respect to the \(x\)-axis, the shell method with respect to the \(x\)-axis, and the general slicing method with slices perpendicular to the \(xy\)-plane and parallel to the \(x\)-axis. Answer: The volume of a hemisphere of radius \(r\) is \(\frac{2}{3}\pi r^3\).

Step 1: Set up the problem

We want to find the volume of a hemisphere of radius \(r\) using three different methods. First, let's find the intersection points of the circle with the coordinate axes. This will help us determine the limits of integration. The circle is given by \(x^2 + y^2 = r^2\) and intersects the \(x\)-axis when \(y=0\), giving \(x=\pm r\) and the \(y\)-axis when \(x=0\), giving \(y=\pm r\). Since we are considering the first quadrant, we will have the points \((r,0)\) and \((0,r)\) as the limits.

Step 2: Volume using the Disk Method

We can find the volume of the hemisphere by revolving the region \(R\) about the \(x\)-axis using the disk method. The formula for the disk of radius \(y\) and thickness \(dx\) is given by \(A(x)=\pi y^2\). Consequently, the differential volume is given by \(dV=A(x)dx=\pi y^2dx\). Integrating, we find the volume: \(\displaystyle V = \int_{0}^{r} \pi y^2 dx\) Now we need to express \(y\) in terms of \(x\). We have \(y^2 = r^2 - x^2\), so \(y = \sqrt{r^2-x^2}\). Substituting into the integral and computing, we have: \(\displaystyle V = \int_{0}^{r} \pi (r^2 - x^2) dx\) \(\displaystyle V = \pi\int_{0}^{r} (r^2 - x^2) dx\) \(\displaystyle V = \pi\left[r^2x - \frac{1}{3}x^3 \right]_{0}^{r}\) \(\displaystyle V = \frac{2}{3}\pi r^3\)

Step 3: Volume using the Shell Method

Now, we'll find the volume of the hemisphere by revolving the region \(R\) about the \(x\)-axis using the shell method. The differential volume is given by \(dV = 2\pi y(x)(dy)\). As the shell forms, we will determine the volume by integrating respect to \(y\): \(\displaystyle V = \int_{0}^{r} 2\pi y x dy\). Now we need to express \(x\) in terms of \(y\). We have \(x^2 = r^2 - y^2\), so \(x = \sqrt{r^2-y^2}\). Substituting into the integral and computing, we have: \(\displaystyle V = \int_{0}^{r} 2\pi y (\sqrt{r^2-y^2}) dy\) \(\displaystyle V = 2\pi \int_{0}^{r} y(\sqrt{r^2-y^2}) dy\) Using trigonometric substitution, let \(y = r \sin{\theta}\), replacement, we get: \(\displaystyle V = 2\pi \int_{0}^{\frac{\pi}{2}} r^2 \sin{\theta}(r\cos{\theta})(r \cos{\theta} d\theta)\) \(\displaystyle V = 2\pi r^3 \int_{0}^{\frac{\pi}{2}} \sin{\theta}\cos^2{\theta} d\theta\) Now, we compute the integral: \(\displaystyle V = 2\pi r^3 \left[\frac{1}{3}\sin^3{\theta} \right]_{0}^{\frac{\pi}{2}}\) \(\displaystyle V = \frac{2}{3}\pi r^3\)

Step 4: Volume using the General Slicing Method

Finally, let's use the general slicing method to find the volume of the hemisphere with slices perpendicular to the \(xy\)-plane and parallel to the \(x\)-axis. The differential volume is given by \(dV=A(z)dz=\frac{1}{2}\pi x^2 dz\). From the equation of the circle, we have \(x^2+z^2=r^2\) and thus \(x^2=r^2-z^2\). Integrating, we find the volume: \(\displaystyle V = \int_{0}^{r} \frac{1}{2}\pi (r^2-z^2) dz\) \(\displaystyle V = \frac{1}{2}\pi\int_{0}^{r} (r^2 - z^2) dz\) \(\displaystyle V = \frac{1}{2}\pi\left[r^2z - \frac{1}{3}z^3 \right]_{0}^{r}\) \(\displaystyle V = \frac{2}{3}\pi r^3\) All three methods yield the same result: \(\displaystyle V = \frac{2}{3}\pi r^3\) , which is the volume of a hemisphere of radius \(r\).