I need to find the volume of a solid of revolution formed by rotating the region bounded by these curves: $y=4+x^2,$ $x=0,$ $y=4+x^2,$ and $x=1$ about the y-axis
Here is the graph: https://www.desmos.com/calculator/x7fxnkwuex
This is what I've got so far(using shell method): $$V=\pi\big(\int_0^4dx + 2\int_4^5(4x+x^3)dx\big)$$
Am I on the right track?
It appears as if you set the limits of integration as the $y$ values. This would work, if you were integrating with respect to $y$. That is, if you were using the washer or disk method.
The volume of a solid of revolution rotating about the y-axis, given the method of cylindrical shells, is given by$$V=2\pi \int_a^bxf(x) dx$$
We are integrating with respect to $x$, so our bounds are from $x=0$ to $x=1$. Plugging in for the equation, we get $$V=2\pi \int_0^1 x(4+x^2)dx = 2\pi \int_0^1(4x+x^3)dx$$ Integrating, we get $$V=2\pi \left[2x^2+\frac 14 x^4\right]_0^1=\frac{9\pi} {2}$$