Exact surface area by rotating about the x-axis

Question

I'm trying to find the surface area by revolving this equation around the x-axis from 0 to 3. $$y^2=x+1$$

I get the answer $$\frac \pi 6 (17\sqrt{17}-5\sqrt5)$$

The answer is correct according to Wolframalpha but my book says the answer is

$$\frac \pi 6 (27\sqrt{27}-5\sqrt5)$$

I'm not sure which one to believe

$\sqrt{27} = 3\sqrt{3}$ and i don't think the book would've left it in an unsimplified form -- ilkely a typo — gt6989b, Oct 24 '17 at 20:25
I'm thinking so too, I went through the problem with a classmate and we can't get $\sqrt27$ to appear — Brady Dean, Oct 24 '17 at 20:26

score 0 · Accepted Answer · answered Oct 24 '17 at 22:32

I get: $$\int_0^32\pi yds=2\pi\int_0^3 \sqrt {x+1}\sqrt {1+(y')^2}dx=2\pi\int_0^3 \sqrt {1+x}\sqrt {1+\frac 14(1+x)^{-1}}dx=2\pi(\frac 12)\int_0^3\sqrt {1+x}\sqrt{\frac{4+4x+1}{x+1}}dx =\pi\int_0^3 \sqrt{5+4x}dx=\pi [\frac16 (5+4x)^\frac 32]_0^3=\pi [\frac 16(17)^\frac 32-(5)^\frac 32]=\frac\pi6 (17\sqrt {17}-5\sqrt 5)$$.

Exact surface area by rotating about the x-axis

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