Find the volume of the solid obtained by rotating the region bounded by
$y=5\sin(5x^2)$, $0 \le x \le \sqrt{(\frac\pi5)}$
about the $y$-axis.
I get the wrong answer using the cylindrical shell formula, so I assume I am plugging the values into the wrong place. I ended up with "$-\pi$" and that is the wrong answer.
Can anyone tell me how to solve this please
$$ V = 2\pi\int_{0}^{\sqrt{\pi/5}}5x\sin(5x^2)dx \to \pi\int_{0}^{\pi}\sin(u)du = 2\pi. $$