<s>[INST] Finding volume of solid in one quadrant - divide total volume by 4? 8? 2? I want to find the volume of the solid produced by revolving the region enclosed by $y=4x$ and $y=x^3$ in the first quadrant. The wording about the first quadrant confuses me but here's my work so far:
I know the volume unrestrained by quadrant is:
$$V = \int_a^b \pi(f(x)^2 - g(x)^2) dx$$
Where $f(x) = 4x$ and $g(x) = x^3$. To find $a$ and $b$, I look for the largest and smallest intersection points between the two functions:
$$\begin{align}
f(x) &= g(x) \\
4x &= x^3 \\
0 &= x^3 - 4x \\
&= x(x-2)(x+2) \\
x &\in \{-2, 0, 2\}
\end{align}$$
Plugging all of these into the volume equation above:
$$\begin{align}
V &= \int_a^b \pi (f(x)^2 - g(x)^2)dx \\
&= \int_{-2}^2 \pi ((4x)^2 - (x^3)^2)dx \\
&= \pi \int_{-2}^2 (16x^2 - x^6)dx \\
&= \pi \left( \int_{-2}^2 16x^2 dx - \int_{-2}^2 x^6 dx \right) \\
&= \pi \left( \left[ \frac{16x^3}{3} \right]_{-2}^2 - \left[ \frac{x^7}{7} \right]_{-2}^2 \right) \\
&= \pi \left( \frac{16(2)^3}{3} - \frac{16(-2)^3}{3} - \frac{(2)^7}{7} - \frac{(-2)^7}{7} \right) \\
&= \pi \left( \frac{128}{3} + \frac{128}{3} - \frac{128}{7} + \frac{128}{7} \right) \\
&= \pi \left( \frac{256}{3} - \frac{256}{7} \right) \\
&= \pi \left( \frac{1024}{21} \right) \\
&= \frac{1024\pi}{21}
\end{align}$$
This is the volume for the entire function. I make an assumption that since I only want one quadrant and the function is symmetric about both the x- and y-axes, I simply divide it by four.
$$\begin{align}
V_{\text{whole}} &= \frac{1024\pi}{21} \\
V_{\text{one quadrant}} &= \frac{1024\pi}{21} \times \frac{1}{4} \\
&= \frac{256\pi}{21}
\end{align}$$
I have no way of verifying my results. Can my assumption be made, or there's a differing method I should be using here? 
If I'm now working in 3D space, would I instead divide it by eight? But if I'm revolving around $x=0$, wouldn't the solid of revolution take four quadrants in 3D space, thus I should divide the total volume by 2?
 [/INST] The first quadrant is defined by $\{(x,y)\in \mathbb R^2 | x,y\ge 0\}$. The $\ge$ instead of $>$ is debatable, but is of no consequence for this problem.
Thus your intersection points are $(0,0)$ and $(2,2)$. So, the volume integral would be $\int_0^2$.
To answer your other question, it would be twice the volume of the correct volume, since you are integrating an even function ($f^2-g^2$) from $-2$ to $2$. Intuitively you are doubling the volume by calculating the volume of the identical rotated ($\pi$ radians) region between $-2$ and $0$.
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