I have two curves: $y_{1} = x^{2} - 4$, $y_{2} = 8 - 2x^{2}$. I have a region bounded by $y_{1}, y_{2}$, $y$-axis, and $x=3$. What is the volume of solid revolve around the $x$-axis?
Attempt:
In the image below, blue is $y_{2}$, red is $y_{1}$. And we see that the region consist of negative values of $y$, so how to calculate the volume using the disc method?
My analysis is that we cannot directly use the formula. For $0 \le x \le 2$, the volume is strictly $$ V_{1} = \pi \int_{0}^{2} (y_{2})^{2} dx$$ because $|y_{2}| \ge |y_{1}|$ in the region. Similarly, $$ v_{2} = \pi \int_{2}^{3} (y_{2})^{2} dx $$ so we have $$ V = \pi \int_{0}^{3} (8-2x^{2})^{2} dx = \pi \int_{0}^{3} 64 -32 x^{2} + 4x^{4} dx$$ $$ = \pi (64x - (32/3)x^{3} + (4/5)x^{5})|_{0}^{3} = 98.4 \pi $$
The 2nd image below is the plot of $|y_{2}|, |y_{1}|$:
