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 region to revolve

   The region is enclosed by the curve $x = \sqrt{5}y^2$ and the lines $y = 1$, $y = -1$, and $x = 0$. This gives us the bounds for $y$ from $-1$ to $1$.

2. Set up the volume integral using the washer method

   When revolving around the y-axis, the volume can be calculated using the formula: $$ V = \pi \int_{a}^{b} \left(R(y)^2 - r(y)^2\right) dy $$ Here, $R(y)$ is the outer radius and $r(y)$ is the inner radius. Here, $R(y) = \sqrt{5}y^2$ and $r(y) = 0$ (since the left boundary is $x=0$).

3. Determine the limits of integration

   The limits of integration are from $y = -1$ to $y = 1$.

4. Set up the integral

   Substituting the functions into the volume integral gives: $$ V = \pi \int_{-1}^{1} \left(\sqrt{5}y^2\right)^2 dy $$

5. Simplify the integrand

   Simplifying the integrand: $$ = \pi \int_{-1}^{1} 5y^4 , dy $$

6. Evaluate the integral

   Calculate the integral: $$ V = 5\pi \left[\frac{y^5}{5}\right]_{-1}^{1} = 5\pi \left(\frac{1^5}{5} - \frac{(-1)^5}{5}\right) $$ This simplifies to: $$ = \pi (1 - (-1)) = 2\pi $$