I have a homework question that asks the following:
Use the shell method to find the volume of the solid generated by revolving the region bounded by the line $y=3x+4$ and the parabola $y=x^2$
about the x-axis.
So I set the following variables:
radius = y
height = $\sqrt{y}-\frac{y}{3}+\frac{4}{3}$
And I end up with the following integral:
$$2\pi\int_0^{16}\left(y^\frac{3}{2}-\frac{y^2}{3}+\frac{4y}{3}\right)dy = 2\pi\left(\frac{2}{5}y^\frac{5}{2}-\frac{y^3}{9}+\frac{2y^2}{3}\right)$$
When I evaluate that over the specified interval, I get: $$\frac{11264\pi}{45}$$
However, my homework marks it incorrect and says the correct answer is $250\pi$. Where am I going wrong? My answer is almost exactly 250, so I am not sure if I am getting it wrong or if my homework is rounding it off.
Best Answer
Start by drawing a picture of the region, then calculate the intersection points. You will see that you need to split the integral into two parts, where the formula for the "height" is different. $$x^2=3x+4$$ The solutions are $x_1=-1$ and $x_2=4$, with the corresponding $y$ coordinates being $1$ and $16$. The shell in this case is horizontal. Below $y=1$ the "height" of the shell is $2\sqrt y$, above it's your formula