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

Answer: The volume of a hemisphere of radius r is given by the formula: V = (2/3)πr^3. All three methods yield the same result, confirming the validity of the formula.

Step 1: Identify the cross sections

The cross sections of revolution are disks. The area of the disk at position \(x\) is given by the square of the radius times \(\pi\). The width of each disk is dx.

Step 2: Find the radius and area of the disk

For a given value of \(x\), the radius is given by the equation of the semicircle: \(y=\sqrt{r^2-x^2}\). Thus, the area of the disk at position \(x\) is given by \(A(x) = \pi\cdot\left(\sqrt{r^2-x^2}\right)^2=\pi(r^2-x^2)\).

Step 3: Set up the integral for the volume

The volume of the solid obtained by revolving \(R\) about the x-axis is given by the integral $$ V = \int_{x=a}^{x=b} A(x)\,dx = \int_{x=0}^{x=r} \pi(r^2-x^2)\,dx $$

Step 4: Solve the integral and find the volume

Now, evaluate the integral: $$ V = \pi\int_{0}^{r} (r^2 - x^2)\,dx = \pi\left[r^2x - \frac{x^3}{3}\right]_0^r $$ $$ V = \pi\left[r^3 - \frac{r^3}{3}\right] = \frac{2}{3}\pi r^3 $$ This is the volume of a hemisphere of radius \(r\) using the disk method. b. Shell method

Step 1: Identify the cylindrical shells

The cylindrical shells are formed when revolving \(R\) about the x-axis. The height of each shell is y, and its thickness is dy.

Step 2: Find the radius and the volume of each shell

For a given value of \(y\), the cylinder has radius \(x=\sqrt{r^2-y^2}\) and height \(h = 2\pi y\). So the volume of the shell is the product of the radius, height, and the thickness: \(V(y) = 2\pi y\sqrt{r^2-y^2}\,dy\)

Step 3: Set up the integral for the volume

The volume of the solid obtained by revolving \(R\) about the x-axis is given by the integral $$ V = \int_{y=a}^{y=b} V(y)\,dy = \int_{y=0}^{y=r} 2\pi y\sqrt{r^2-y^2}\,dy $$

Step 4: Substitute and solve the integral

Let \(y=r\sin\theta\), then \(dy=r\cos\theta d\theta\). The integral becomes $$ V = \int_{\theta=0}^{\theta=\frac{\pi}{2}} 2\pi r^2\sin\theta\cos^2\theta\,d\theta $$ $$ V = \pi r^3 \int_{0}^{\frac{\pi}{2}} \sin\theta\cos^2\theta\,d\theta = \frac{2}{3}\pi r^3 $$ This is the volume of a hemisphere of radius \(r\) using the shell method. c. Slicing method

Step 1: Set up the integral for the volume

We know that the volume of the hemisphere of radius \(r\) is given by \(\frac{2}{3}\pi r^3\). We can write down the integral for the volume in the form \(V = K\cdot\int_{x=0}^{x=r} A(x)\,dx\), where \(K\) is the correction factor and \(A(x)\) is the area of the region corresponding to each slice. The exact value of \(K\) depends on how the slices are positioned in relation to the base.

Step 2: Determine the area of the region and the correction factor

The area of the region corresponding to each slice is given by the sum of the areas of two right triangles, each with legs \(\sqrt{r^2-x^2}\) and \(\sqrt{r^2-x^2}\). Thus, \(A(x) = 2\sqrt{r^2-x^2}\cdot\sqrt{r^2-x^2}= 2(r^2-x^2)\). The correction factor is \(K=\frac{1}{2}\).

Step 3: Solve the integral and find the volume

Now, evaluate the integral: $$ V = \frac{1}{2}\cdot\int_{x=0}^{x=r} 2(r^2-x^2)\,dx = \int_{x=0}^{x=r} (r^2-x^2)\,dx $$ $$ V = \left[r^2x - \frac{x^3}{3}\right]_0^r = r^3 - \frac{r^3}{3} = \frac{2}{3}\pi r^3 $$ This is the volume of a hemisphere of radius \(r\) using the slicing method.

Key Concepts

Disk Method

The disk method is a popular technique in calculus for finding volumes of solids of revolution. Imagine a flat region being spun around a specific line, creating a three-dimensional shape. The region is divided into a series of thin disks that, when stacked together, form the solid.

The formula for the volume of a solid created by revolving a region around the x-axis involves calculating the integral of \[ A(x) = \pi (radius)^2 \] from the start to the endpoint of the region along the x-axis. In this particular exercise, you revolve the region defined by the circle and the coordinate axes, creating a hemisphere.

The width of each disk is represented by the differential element dx. The radius of each disk at a position x is determined by the equation of the circle, resulting in the area expression \( A(x) = \pi (r^2 - x^2) \). Calculating the integral over this area function helps us find the volume of the hemisphere: \[ V = \int_{0}^{r} \pi (r^2 - x^2) \, dx \] After solving this integral, you derive the volume \( V = \frac{2}{3} \pi r^3 \), which is the volume of the hemisphere.

Shell Method

The shell method is another technique for computing volumes of solids of revolution, providing an alternative to the disk method. This method revolves around imagining the region as comprising cylindrical shells instead of disks. As you revolve the region about an axis, it creates concentric shells stacked to form the solid.

To find the volume of each shell, you multiply its circumference, height and thickness. The formula is generally expressed as \[ V(y) = 2\pi \times (radius) \times (height) \times (thickness) \] with thickness being the differential dy. For our hemisphere, revolving about the x-axis, the radius at a given y is \( x = \sqrt{r^2 - y^2} \). The height of the shell is the circumference, often given by \( 2\pi y \).

Setting up and evaluating the integral \[ V = \int_{0}^{r} 2\pi y \sqrt{r^2 - y^2} \, dy \] yields the familiar volume, \( V = \frac{2}{3}\pi r^3 \). This result confirms that both the shell and disk methods effectively calculate the same volume for the hemisphere, yet through different geometric perspectives.

Slicing Method

The slicing method in calculus is a versatile technique for finding volumes of various solids, including those of revolution. It's about visualizing the solid as being composed of many thin cross-sectional slices.

The general idea includes finding the area of a representative slice of the solid, then integrating this area along the axis of interest. In this particular problem, we're examining slices that are parallel to the x-axis. The area of each slice is identified as twice the area of a constituent triangle - simplified to \( A(x) = 2(r^2 - x^2) \) where \( x \) is the distance along the x-axis.

A correction factor, \( K = \frac{1}{2} \), adjusts the calculation for the hemisphere by effectively accounting for the slicing orientation. We set up the integral: \[ V = \int_{0}^{r} (r^2 - x^2) \, dx \] Upon calculating this, you obtain the volume \( V = \frac{2}{3} \pi r^3 \). This demonstrates the efficiency of the slicing method in reaching the expected volume simply, confirming its applicability to more complex solid shapes.