Question

In Exercises 39-42, find the volume of the solid analytically. The solid lies between planes perpendicular to the \(y\) -axis at \(y=0\) and \(y=2 .\) The cross sections perpendicular to the \(y\) -axis are circular disks with diameters running from the \(y\) -axis to the parabola \(x=\sqrt{5} y^{2}\)

Short answer

The volume of the solid is \(\frac{32\pi}{4}\) cubic units.

Step-by-step solution

Define the Radius Function

Recognize that the diameter of the disk is the \(x\)-value given by the parabola, which is \(x = \sqrt{5}y^{2}\). Therefore, the radius \(f(y)\) of the disk will be half of this diameter, so \(f(y) = \frac{\sqrt{5}y^{2}}{2}\).

Set Limits of Integration

The planes at \(y = 0\) and \(y = 2\) give us the limits of integration. Therefore, the limits of the integral are \(a = 0\) and \(b = 2\).

Integrate to Find Volume

Substitute the radius function and the limits into the volume integral to find the volume of the solid. This will result in \(\int_0^2 \pi (\frac{\sqrt{5}y^{2}}{2})^2 dy\) which simplifies to \(\int_0^2 \frac{5\pi }{4} y^4 dy\). Solve the integral to obtain the volume of the solid.

Calculate Definite Integral

Solve the definite integral \(\int_0^2\frac{5\pi }{4}y^4 dy\). The antiderivative of \(y^4\) is \(\frac{1}{5}y^5\). Therefore, the definite integral is: \(\frac{5\pi }{4} *\frac{1}{5}y^5\Bigg|_0^2 = \frac{5\pi}{4} * \frac{1}{5}*(32 - 0) = \frac{32\pi}{4}\).