$y=x^2$ and $y=12 - 2x^2$ intersect at the points $(-2,4)$ and $(2,4)$. The region bounded by the curves and the $y$-axis is rotated about the $y$-axis. By splitting the shading region into two parts, or otherwise, find the volume of the solid formed.
When I draw it, the shaded area is on right side of the y-axis and above the x-axis)
I got $24\pi$ but I'm certain its wrong please show full working.
Firstly draw a clear diagram.
The volume of revolution around the y-axis is given by:
$$\pi\int_{y_1}^{y_2}x^2dy$$
Your curve has two distinct areas so work out each separately:
$$\pi\left(\int_0^4x_1^2dy+\int_4^{12}x_2^2dy\right)$$
Note that the limits used are the y-values as that is the axes we are integrating around.
Next rearrange the equations into the form $x^2=...$ and substitute in:
$$\pi\left(\int_0^4ydy+\int_4^{12}\left(6-\frac{y}{2}\right)dy\right)$$
$$=\pi\left(\left[\frac{y^2}{2}\right]_0^4+\left[6y-\frac{y^2}{4}\right]_4^{12}\right)$$
$$=\pi\left(\frac{4^2}{2}-0+6\cdot12-\frac{12^2}{4}-\left(6\cdot4-\frac{4^2}{4}\right)\right)$$
$$=24\pi\space units^3$$