91Ó°ÊÓ

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

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 $$ V = \int_0^2 \pi (y^2)^2 dy $$

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 \(\frac{32\pi}{5}\).