Question

True or False If the region enclosed by the \(y\) -axis, the line \(y=2,\) and the curve \(y=\sqrt{x}\) is revolved about the \(y\) -axis, the volume of the solid is given by the definite integral \(\int_{0}^{2} \pi y^{2} d y\) Justify your answer.

Short answer

False. The correct volume of the solid is given by the definite integral \( \int_{0}^{2} \pi y^{4} \,dy\), not \( \int_{0}^{2} \pi y^{2} \,dy\).

Step-by-step solution

Understanding the Disk Method

The disk method in calculus is used to compute the volume of the solid generated by revolving a region around a line. Here, the radius of the disks will be \(x\), the distance from the y-axis, the height will be \(dy\), and thus, the volume of each disk is \( \pi [x(y)]^{2} \,dy\). Because the radius is a function of \(y\), this integral in terms of \(y\) will give the volume of the solid.

Y as a Function of X

It's stated that \(y=\sqrt{x}\), so expressing \(x\) as a function of \(y\), we find that \(x = y^2\). This is necessary because we're revolving around the y-axis.

Substituting to Confirm the Equation

Next, we substitute this in the disk method equation: \( \pi [x(y)]^{2} \,dy = \int \pi [y^{2}]^2 dy\). This simplifies to \( \int_{0}^{2} \pi y^{4} dy\).

Result Synopsis

Clearly, the integral found in step 3, \( \int_{0}^{2} \pi y^{4} dy\), does not match the integral in the problem, \( \int_{0}^{2} \pi y^{2} dy\). Thus the statement in the problem is false.