Volume between two functions rotated about the x-axis I'm asked to find the volume of the shape that emerges when the curve $y = 14-x^2$  (above $y = 5$) is rotated about the x-axis. 
I simply put $14-x^2 = 5$ and got $x=3$ or $x=-3$
From y = 5 we also obtain $f(x) = x^2-9$
So now I want to find $\pi \int_{0}^{3} (x^2-9)^2$ and multiply this by 2 to get the whole volume. I get the volume $\frac{1296\pi}{5}$ from this though, which is incorrect according to my solutions manual. What am I doing wrong?
 A: The region you are interested in is shown below:

When you rotate that around the $x$-axis, you get a bunch of discs. Can you figure out what the outer and inner radius of each disc is? That should help you develop the proper integral.
Your use of $x^2-9$ as the integrand would be correct if this region rested on the $x$-axis. But the volume you get by rotating the figure above and the volume you get when you rotate a similar region resting on the $x$-axis are not the same.
A: If you want to use this $$2\pi\int_0^3(9 - x^2)^2dx$$, you need to lift up $x$-axis from $y = 0$ to $y = 5$. Then the areas of the circle (rotated around $x$-axis at y = 5) is the same as integral of the circle of radius $3$ (rotated around the $x$-axis at $y = 0$). However, this time we are asked to rotate $(14 -x^2)$ around $x$-axis$(y = 0)$. We need to subtract inner area(the circle of radius $5$) from total area(the circle of radius $(14 - x^2)$) Therefore, $2\pi\int_0^3(14 - x^2)^2dx$(Total donut) $-$ $2\pi\int_0^3(5)^2dx$(Inner donut). The answer is $\frac{3096}{5}\pi$
A: Well, when you rotate a constant around the x-axis we get:
$$\mathscr{V}_{\space\text{n}_1}\left(\text{a},\text{b}\right):=\int_\text{a}^\text{b}\pi\cdot\text{n}_1^2\space\text{d}x=\pi\cdot\text{n}^2\cdot\left[x\right]_\text{a}^\text{b}=\pi\cdot\text{n}_1^2\cdot\left(\text{b}-\text{a}\right)\tag1$$
And for a quadratic equation we get:
$$\mathscr{V}_{\space\text{n}_2,\text{n}_3,\text{n}_4}\left(\text{a},\text{b}\right):=\int_\text{a}^\text{b}\pi\cdot\left(\text{n}_2\cdot x^2+\text{n}_3\cdot x+\text{n}_4\right)^2\space\text{d}x\tag2$$
Now, when we have:
$$-\frac{1}{2}\cdot\frac{\text{n}_3}{\text{n}_2}>\text{n}_1\tag3$$
Than we know that:
$$\text{n}_1=\text{n}_2\cdot x^2+\text{n}_3\cdot x+\text{n}_4\space\Longleftrightarrow\space x_{\pm}=\frac{-\text{n}_3\pm\sqrt{\text{n}_3^2-4\cdot\text{n}_2\cdot\left(\text{n}_4-\text{n}_1\right)}}{2\cdot\text{n}_2}\tag4$$
So:


*

*$$\text{a}=x_-=\frac{-\text{n}_3-\sqrt{\text{n}_3^2-4\cdot\text{n}_2\cdot\left(\text{n}_4-\text{n}_1\right)}}{2\cdot\text{n}_2}\tag5$$

*$$\text{b}=x_+=\frac{-\text{n}_3+\sqrt{\text{n}_3^2-4\cdot\text{n}_2\cdot\left(\text{n}_4-\text{n}_1\right)}}{2\cdot\text{n}_2}\tag6$$


For the final volume we get:
$$\mathscr{V}_{\space\text{T}}=\mathscr{V}_{\space\text{n}_2,\text{n}_3,\text{n}_4}\left(x_-,x_+\right)-\mathscr{V}_{\space\text{n}_1}\left(\text{a},\text{b}\right)=$$
$$\int_{x_-}^{x_+}\pi\cdot\left(\text{n}_2\cdot x^2+\text{n}_3\cdot x+\text{n}_4\right)^2\space\text{d}x-\int_{x_-}^{x_+}\pi\cdot\text{n}_1^2\space\text{d}x=$$
$$\pi\cdot\int_{x_-}^{x_+}\left(\left(\text{n}_2\cdot x^2+\text{n}_3\cdot x+\text{n}_4\right)^2-\text{n}_1^2\right)\space\text{d}x\tag7$$

When $\text{n}_2=-1$ and $\text{n}_3=0$, we can simplify $\left(7\right)$, to:
$$\mathscr{V}_{\space\text{T}}=\pi\cdot\int_{-\sqrt{\text{n}_4-\text{n}_1}}^\sqrt{\text{n}_4-\text{n}_1}\left(\left(\text{n}_4-x^2\right)^2-\text{n}_1^2\right)\space\text{d}x=$$
$$\frac{8\pi}{15}\cdot\left(\text{n}_4-\text{n}_1\right)^\frac{3}{2}\cdot\left(3\cdot\text{n}_1+2\cdot\text{n}_4\right)\tag8$$
