Question

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.
y = 5x2, y = 30x − 10x2

Answer 1

I got 25pi/2 but that seems to be wrong.

Answer 2

\[\int\limits_{0}^{1}2pix (30x-10x^2-5x^2)dx\]

Answer 3

I did that and then got the final answer as 25pi/2. But I didnt get the right answer

Answer 4

Looks like the integral setup needs work.

I find drawing the region with at least one shell to be very helpful. To make it easy to know where to plot the second parabola, you can complete the square and find its vertex:
\[\begin{align*}
30x-10x^2&=-10(x^2-3x)\\[1ex]
&=-10\left(x^2-3x+\frac94-\frac94\right)\\[1ex]
&=-10\left(\left(x-\frac32\right)^2-\frac94\right)\\[1ex]
&=\frac{45}2-10\left(x-\frac32\right)^2
\end{align*}\]so we know this is a parabola opening downward with its vertex at \(\left(\dfrac32,\dfrac{45}2\right)\). Meanwhile, the parabola \(5x^2\) opens upward and has its vertex at the origin.

Find their intersections, if any:
\[\begin{align*}
5x^2&=30x-10x^2\\[1ex]
15x^2-30x&=0\\[1ex]
x^2-2x&=0\\[1ex]
x(x-2)&=0
\end{align*}\]so the parabolas meet at \(x=0\) and \(x=2\), which conflicts with your current setup with the integral ranging from \(0\) to \(1\).

Now here's a rough sketch of the region: