I'm a little confused about how to properly apply the integrals used to calculate the area of a surface of revolution.
Find the exact area of the surface obtained by rotating the curve about the x-axis: $y^2 = x + 1, 0 \le x \le 3$
Rewriting the equation, I have $y= \sqrt{x+1}$. Taking its derivative yields $\frac {dy}{dx}=\frac{1}{2\sqrt{x+1}}$. Substituting these values into what I believe is the appropriate formula, I have (some steps in the algebra omitted for conciseness):
$$S = \int_0^32\pi y\sqrt{1+(\frac{1}{2\sqrt{x+1}})^2} dx = 2\pi\int_0^3\sqrt{x+1}\sqrt{1+\frac{1}{4(x+1)}} dx = 2\pi\int_0^3\sqrt{\frac{4x+5}{4}}$$
Which eventually simplifies to $\pi\int_0^3\sqrt{4x+5}dx$. Applying u-substitution then gets me
$$\frac{\pi}{4}\int_5^{17}(u)^\frac{1}{2}du = \frac{\pi}{6}(17\sqrt{17} - 5\sqrt{5})$$
Which differs from my textbook's (Stewart's Calculus, 8th ed.) answer:
$$\frac{\pi}{6}(27\sqrt{27} - 5\sqrt{5})$$
What did I do wrong? When I saw that my answer was the same except for that $27\sqrt{27}$, I assumed I'd made some mistake in my algebra somewhere, but I've looked over my work three times now and I can't seem to find any.