Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about the y-axis.
$y=x^2$, $y = 5 -4x$, $x = 0$ for $x \ge0$
I have the integral set up as
$$2\pi \int_{-5}^0(x)(-4x+5-x^2) \, dx$$
I'm having issues typing that in correctly. The integral evaluates to $875\pi/5$ when the answer is $11\pi/6$. Why is my setup wrong? Everything I look at about it seems right and am having trouble.
The region that is being rotated can be seen below; it's the region below $y=5-4x$ and above $y=x^2$, with $x\in[0,1]$. So, the volume is$$\int_0^12\pi x(5-4x-x^2)\,\mathrm dx=\frac{11\pi}6.$$