<s>[INST] Volume of Region Bounded by $y = \sqrt{x}$ and $y = 18-x^2$ I haven't a clue why I'm having so much issues solving this problem. The two curves are $y=\sqrt{x}$ and $y=18-x^2, x \ge 1$, and rotated about the $y$-axis.
If I'm not mistaken, I need to use the shell method. In that case, the volume integration formula is $\int 2\pi x f(x)dx$.
So my thickness is $dx$, my height is $f(x)$, I believe my range is from $1$ to $\sqrt{18}$ ($1$ because $x \ge 1$ and $\sqrt{18}$ because $0 = 18-x^2$).
So I integrate $2\pi \cdot x \cdot x^{1/2} dx$ and get $(2/5)x^{5/2}$ from $1$ to $\sqrt{18}$, but that isn't even close to the answer.
I've been working on this problem far too long and am getting brain fog.
 [/INST] Note that the graph of $y = \sqrt{x}$ and $y = 18-x^2$ shows that these two curves meet at $x = 4$ and $y = 18-4^2 = \sqrt{4} = 2$.  Write out explicitly $f$ in the question body.
$$f(x) = 18-x^2-\sqrt{x} \quad \forall\,x \in [1,4]$$
\begin{align}
\text{Required volume of revolution}
&= 2\pi \int_1^4 x f(x) dx \\
&= 2\pi \int_1^4 (18x - x^{3/2} - x^3) dx \\
&= 2\pi \left[9x^2 - \frac{x^{5/2}}{5/2} - \frac{x^4}{4} \right]^4_1 \\
&= 2\pi \left[ 9(4^2 - 1^2) - \frac{64}{5} + \frac25 - \frac{4^4 - 1^4}{4} \right] \\
&= \frac{1177\pi}{10}
\end{align}
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