Find the exact area of the surface obtained by rotating the curve about the x-axis. y=sqrt(1+4x) 1

Question

Find the exact area of the surface obtained by rotating the curve about the x-axis. y=sqrt(1+4x) 1<x<5

anonymous · Answer

Are you allowed to use Calculus?

anonymous · Answer

Yes.

anonymous · Answer

That's what I'm trying to do. I got up to $$2\pi \int\limits_{1}^{5} \sqrt{1+4x} \sqrt{1/\sqrt{1+4x}}$$

anonymous · Answer

That formula looks a bit confusing to me CatLove9, try to think what the solid looks like, the radius especially.

dumbcow · Answer

the 2nd radical isn't quite right
for surface area integral, multiply by dA
$$dA = \sqrt{1+f'(x)^{2}}$$
$$f'(x) = \frac{2}{\sqrt{4x+1}}$$
$$\rightarrow \sqrt{1+\frac{4}{4x+1}} = \sqrt{\frac{4x+5}{4x+1}}$$

anonymous · Answer

thought it would be $$1\over2\sqrt{4x+1}$$

dumbcow · Answer

chain rule...multiply that by derivative of inside

anonymous · Answer

Oh, yes yes, I forgot about that.

anonymous · Answer

How did you get the 4x on the top again? I forgot how to do that.

dumbcow · Answer

combining fractions
change the 1 to 4x+1/4x+1

anonymous · Answer

Okay I got you. so then what happens after that.

dumbcow · Answer

you have to integrate...it comes out nicely because the sqrt(1+4x) cancels out

$$2\pi \int\limits_{1}^{5}\sqrt{1+4x}*\frac{\sqrt{4x+5}}{\sqrt{1+4x}} dx$$

anonymous · Answer

I see, thank you!

dumbcow · Answer

your welcome...for an answer i got 98pi/3