Problem 38 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the disk or washer method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$y=\sqrt{x}, y=0, x=4$$

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\frac{32\pi}{5}\).

Step 1: Understand the given curves and the region R

The region R is given by the curves \(y=\sqrt{x}, y=0, x=4\). We can understand the region R as follows: - The curve \(y=\sqrt{x}\) is the upper boundary of the region R - The line \(y=0\) is the lower boundary of the region R, which is also the x-axis - The line \(x=4\) is the vertical boundary of the region R on the right side

Step 2: Find the volume using the disk method

We can use the disk method to find the volume of the solid generated when the region R is revolved about the y-axis. The disk method involves setting up an integral representing the sum of the volumes of small disks that are formed when the region R is revolved about the y-axis. The thickness of each disk will be \(dx\), and their radius will be given by the distance from the curve \(y\) to the vertical line \(x=0\). So the radius is equal to the value of x. From the curve \(y=\sqrt{x}\), we get \(x=y^2\). The volume of each small disk can be given by \(dV = \pi r^2 dx\). So, the volume of the solid can be given as $$ V = \int_a^b \pi r^2 dx $$ In our case, \(r=x=y^2\), so the integral becomes $$ V = \int_a^b \pi (y^2)^2 dy $$ We need to find the limits of integration \(a\) and \(b\). To find the y-coordinate where the curve \(y=\sqrt{x}\) intersects the line \(x=4\), we set \(x=4\), $$ y = \sqrt{4} = 2 $$ So, the limits of integration are \(a=0\) (lower boundary) and \(b=2\) (intersection point). Now, the integral becomes $$ V = \int_0^2 \pi (y^2)^2 dy $$

Step 3: Evaluate the integral

Now, we just have to evaluate the integral: $$ V = \int_0^2 \pi (y^2)^2 dy $$ The integral can be simplified as $$ V = \pi\int_0^2 y^4 dy $$ Now, integrate with respect to y, $$ V = \pi\left[\frac{1}{5}y^5\right]_0^2 $$ $$ V = \pi\left(\frac{1}{5} \cdot 2^5 - \frac{1}{5} \cdot 0^5 \right) $$ $$ V = \pi\left(\frac{1}{5} \cdot 32\right) $$ $$ V = \frac{32\pi}{5} $$ So, the volume of the solid generated when region R is revolved about the y-axis is \(\frac{32\pi}{5}\).