I want to compute the volume obtained by rotating the bounded region of $y=3-x^2$, $y=2x$, $x \leq 0$ around the $y$-axis. I want to use the cylindrical shell method, so my integral is:
$\displaystyle V=2\pi\int_{-3}^0 x(3-x^2-2x) \,\mathrm{d}x$
However, my answer is $\displaystyle -\frac{45\pi}{2}$. It is a negative number. I don't think it is correct because when I try to compute the area between $y=0$ and $y=x^2-4$, I still have a positive number even though the curve is under the $x-axis$, so I think there must be $-x$ or reversing the limit of integration.
What would be the correct way to set up this integral without using the absolute value.
Thank you.
PS: This is relating to section 5.2 and 5.3 in Stewart Calculus book.