The region bound by $y=x^2, y=-x, y=3$, rotated about $x$-axis. I tried to solve it by first finding the volume of the region bounded by $y=x^2$ and $y=3$ ($\frac{36\sqrt{3}}{5}\pi$). Then I found the volume of the region bound by $y=-x$ and $y=x^2$ ($\frac{2}{15}\pi$). Then I subtracted the latter from the former and got the answer of around $38.76$.
I wanted to know whether my answer is correct and if there is another way of solving it.
I think the region to be revolved about $x$-axis is a triangle-like region determined by the intersection points $A(-3,3)$, $B(-\sqrt3,3)$ and $C(-1,1)$. On that region, above the interval $[-3,-\sqrt3]$, the upper function is $y=3$ and the lower function is $y=-x$; above the interval $[-\sqrt3,-1]$, the upper function is $y=x^2$ and the lower function is $y=-x$. Hence,
$$V=\pi\int_{-3}^{-\sqrt3}(3^2-(-x)^2)dx+\pi\int_{-\sqrt3}^{-1}((x^2)^2-(-x)^2)dx=\frac{272-108\sqrt3}{15}\pi$$
By WA.