# Thread: Area of Surface of Revolution

1. ## Area of Surface of Revolution

Hey, I have an area problem that looks like it should simplify, but the integral I have set up does not seem to be solvable with my current skill set. If anyone has a second to look over my set up to see where I went wrong, I would be extremely appreciative!

Compute the area of the surface generated by revolving the curve $y=tan(x)$ , where $x \in [0,\frac{\pi}{4}]$, about the x-axis
So using the SA formula: $r = y = tan(x)$ and $\frac{dy}{dx}=sec^2(x)$.

So $\sqrt{1+(\frac{dy}{dx})^2} = \sqrt{1+sec^4(x)}$.

This gives rise to the SA integral:

$S = 2\pi \int_{0}^{\frac{\pi}{4}} tan(x) \sqrt{1+sec^4(x)} dx$

I cannot figure out how to solve this integral, any hints? Just a hint as well please, not looking for a complete solution. Thanks

2. Originally Posted by Kasper
Hey, I have an area problem that looks like it should simplify, but the integral I have set up does not seem to be solvable with my current skill set. If anyone has a second to look over my set up to see where I went wrong, I would be extremely appreciative!

So using the SA formula: $r = y = tan(x)$ and $\frac{dy}{dx}=sec^2(x)$.

So $\sqrt{1+(\frac{dy}{dx})^2} = \sqrt{1+sec^4(x)}$.

This gives rise to the SA integral:

$S = 2\pi \int_{0}^{\frac{\pi}{4}} tan(x) \sqrt{1+sec^4(x)} dx$

I cannot figure out how to solve this integral, any hints? Just a hint as well please, not looking for a complete solution. Thanks
Try letting

$u = \sqrt{1 + \sec^4x}$.

3. Originally Posted by Danny
Try letting

$u = \sqrt{1 + \sec^4x}$.
Hmm, I never thought to do that!

Does this look correct then following that substitution?

$S=2\pi \int_0^\frac{\pi}{4} tan(x)\sqrt{1+sec^4(x)} dx$

$Let \ u=\sqrt{1+sec^4(x)}$

$du = \frac{4sec^4(x)tan(x)}{2\sqrt{1+sec^4(x)}}dx$

$\frac{2u}{4sec^4(x)}du=tan(x)dx$

And due to the fact $u=\sqrt{1+sec^4(x)}$ : $4sec^4(x) = 4u^2-4$

So $S=2\pi \int_0^\frac{\pi}{4} tan(x)\sqrt{1+sec^4(x)} dx = 2\pi \int_0^\frac{\pi}{4} \frac{2u}{4u^2 -4} du = \pi \int_0^\frac{\pi}{4} \frac{u}{u^2 -1} du$, right? Now I suppose I could do another substitution, integrate and resubstitute back to x to evaluate.

Thanks for helping me with that! I wouldn't have seen that u-sub for a looong time.

4. Originally Posted by Kasper
Hmm, I never thought to do that!

Does this look correct then following that substitution?

$S=2\pi \int_0^\frac{\pi}{4} tan(x)\sqrt{1+sec^4(x)} dx$

$Let \ u=\sqrt{1+sec^4(x)}$

$du = \frac{4sec^4(x)tan(x)}{2\sqrt{1+sec^4(x)}}dx$

$\frac{2u}{4sec^4(x)}du=tan(x)dx$

And due to the fact $u=\sqrt{1+sec^4(x)}$ : $4sec^4(x) = 4u^2-4$

So $S=2\pi \int_0^\frac{\pi}{4} tan(x)\sqrt{1+sec^4(x)} dx = 2\pi \int_0^\frac{\pi}{4} \frac{2u}{4u^2 -4} du = \pi \int_0^\frac{\pi}{4} \frac{u{\color{red}^2}}{u^2 -1} du$, right? Now I suppose I could do another substitution, integrate and resubstitute back to x to evaluate.

Thanks for helping me with that! I wouldn't have seen that u-sub for a looong time.
You should pick up another u (see above in red). Also, switch your limits.

5. Originally Posted by Danny
You should pick up another u (see above in red). Also, switch your limits.
Sorry I'm not clear on where the square is coming from, or why the limits should be switched.

Wouldnt the denominator in the derivative of u be $2(\sqrt{1+sec^4x})^2$ for that to be a $u^2$? I must have missed something.

6. $\int{tan(x)\left(\sqrt{1+Sec^4(x)}\right)}dx=\int{ \frac{2u}{4\left(u^2-1\right)}(u)}du$

7. OH! I forgot about the original $\sqrt{1+sec^4x}$ kicking around there, that makes total sense now. And I suppose by switch your limits, you mean plug that into the u= and find new limits in terms of u. I've heard of doing that, but we've always back-substituted is why I was confused. Thought you meant flip them, haha. Thank you very much!
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