Questions: Find the volume of the solid generated by revolving the region bounded by the given curve and lines about the y-axis. x=3/(sqrt(y+1)), x=0 ; y=0, y=2 V= (Type an exact answer, using pi as needed:)

Solution Steps

To find the volume of the solid generated by revolving the given region about the y-axis, we can use the method of cylindrical shells. The formula for the volume of a solid of revolution using cylindrical shells is:

\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \]

However, since we are revolving around the y-axis, we need to express \(x\) as a function of \(y\). The given curve is \(x = \frac{3}{\sqrt{y+1}}\). The bounds for \(y\) are from 0 to 2. We will integrate with respect to \(y\).

\[ V = 2\pi \int_{0}^{2} x \cdot dy = 2\pi \int_{0}^{2} \frac{3}{\sqrt{y+1}} \, dy \]

To find the volume of the solid generated by revolving the region bounded by the curve \( x = \frac{3}{\sqrt{y+1}} \) and the lines \( x = 0 \), \( y = 0 \), and \( y = 2 \) about the \( y \)-axis, we use the method of cylindrical shells. The volume \( V \) is given by:

\[ V = 2\pi \int_{0}^{2} \frac{3}{\sqrt{y+1}} \, dy \]

We evaluate the integral:

\[ 2\pi \int_{0}^{2} \frac{3}{\sqrt{y+1}} \, dy \]

The integral evaluates to:

\[ 2\pi \left[ 6\sqrt{y+1} \right]_{0}^{2} = 2\pi \left( 6\sqrt{3} - 6 \right) \]

Simplify the expression:

\[ 2\pi \left( 6\sqrt{3} - 6 \right) = 12\pi (\sqrt{3} - 1) \]

The numerical value of the volume is approximately:

\[ 27.5977 \]

Final Answer