Find the volume of the solid generated by revolving the region enclosed by x = √(5y²) and the lines y = 1, x = 0, y = -1 about the y-axis.

Steps to Solve

1. Identify the curves and boundaries

The region is bounded by the curve ( x = \sqrt{5y^2} ), which simplifies to ( x = \sqrt{5} |y| ). The boundaries are ( y = 1 ), ( y = -1 ), and ( x = 0 ).

2. Determine the limits of integration

For this volume calculation, the limits of integration will be from ( y = -1 ) to ( y = 1 ).

3. Set up the volume integral

We will use the formula for the volume of revolution about the y-axis: $$ V = \pi \int_{a}^{b} [f(y)]^2 , dy $$

In this case, ( f(y) = \sqrt{5}|y| ) (the radius of revolution). The volume integral becomes: $$ V = \pi \int_{-1}^{1} ( \sqrt{5} |y| )^2 , dy $$

4. Simplify the integral

We need to square the function: $$ V = \pi \int_{-1}^{1} 5y^2 , dy $$

5. Calculate the integral

Since the function ( y^2 ) is symmetric, we can simplify the integral: $$ V = \pi \int_{0}^{1} 5y^2 , dy \times 2 $$

Calculating the integral: $$ V = 2\pi \cdot 5 \cdot \left[ \frac{y^3}{3} \right]_{0}^{1} = 2\pi \cdot 5 \cdot \frac{1}{3} = \frac{10\pi}{3} $$

6. Final volume

Thus, the volume of the solid generated is: $$ V = \frac{10\pi}{3} $$