Finding the surface area of two solids of revolution I have a couple of questions that are similar in nature:

1) I am trying to find the surface area of this when I rotate it around the x-axis. I have $y = \sqrt{5-x}$ when $3 \leq x \leq 5$
Say $dy/dx = \frac{1}{2}(5-x)^{\frac{-1}{2}}$. Then $(dy/dx)^2 = \frac{1}{4}(5-x)^{-1}$ and thus
$$SA = 2 \pi \int_3^5 \sqrt{5-x} * \sqrt{1 + \frac{1}{4} (5-x)^{-1}}$$
I want to find the exact surface area. Where do I go from here?
EDIT: Can I do this?
$$\begin{align}
SA &= 2 \pi \int_3^5 \sqrt{(5-x) + \frac{1}{4}}dx\\
&= 2 \pi \int_3^5 \sqrt{(5-x) + \frac{1}{4}}dx\\
&= 2 \pi \int_3^5 \sqrt{\frac{21}{4} - x} dx\\
&= 2 \pi \int_3^5 \sqrt{\frac{21}{4} - \frac{4x}{4}} dx\\
&= \pi \int_3^5 \sqrt{21 - 4x} dx
\end{align}$$
Am I on the right track?

2) In this case, I have $y^2 = x + 1$ when $0 \leq x \leq 3$
So $y = \sqrt{x+1}$ and so $$\frac{dy}{dx} = \frac{1}{2}(x+1)^{\frac{-1}{2}}$$ and $$\left( \frac{dy}{dx} \right)^2 = \frac{1}{4}(x+1)^{-1}$$
So
$$SA = 2 \pi \int_0^3 \sqrt{x+1} \sqrt{1 + \frac{1}{4 (x+1)}}$$
EDIT
Am I on the right track:
$$SA = 2 \pi * \int_0^3 \sqrt{x+1 + \frac{1}{4}}$$
$$SA = 2 \pi * \int_0^3 \sqrt{x+\frac{5}{4}}$$
and $u = x + \frac{5}{4}$ and $du = dx$
so
$$2 \pi \int_0^3 \sqrt{x + \frac{5}{4}}dx$$
$$2 \pi \int_{\frac{5}{4}}^{\frac{17}{4}} \sqrt{u}du$$
$$2 \pi \frac{2}{3}u^{\frac{2}{3}}\biggr]_{\frac{5}{4}}^{\frac{17}{4}}$$
$$\frac{4 \pi}{3} \bigg(\frac{17}{4} \sqrt{\frac{17}{4}} - \frac{5}{4} \sqrt{\frac{5}{4}}\bigg)$$

For the first question, notice:
$$\begin{align}
\sqrt{5-x} * \sqrt{1 + \frac{1}{4} (5-x)^{-1}} &= \sqrt{(5-x)\left( 1 + \frac{1}{4} (5-x)^{-1} \right)} \\
&= \sqrt{(5-x)+(1/4)} \\
&= \sqrt{-x+ \frac{21}{4}}
\end{align}$$
Thus the integral becomes
$$SA = 2\pi \int_3^5  \sqrt{-x+ \frac{21}{4}} dx$$
A simple $u$-substitution where $u$ equals what is under the radicand should make the rest of the integral easy to calculate.
EDIT: To address the edit you made to the question OP, yes, that method is also a valid way of approaching this. You will still have to make the $u$-substitution of $u=21-4x$ there, though, so in my opinion just doing it right off the bat results in less computation. Matter of opinion though.

The same principle follows for your second question, yielding
$$SA = 2\pi \int_0^3 \sqrt{x + \frac 5 4} dx$$
And of course, a substitution where $u$ is the radicand makes the rest a fairly simple calculation.

For positive real numbers $a,b$, it is helpful to note that
$$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$$
This is the property we made use of here.
