Find the surface area

**VitaX** · January 20th 2010, 02:07 AM

Find the surface area

$y=Tanx$ limits of $x=0$ and $x=\frac{\pi}{4}$ rotated about the x-axis

$\frac{dy}{dx} = Sec^2x$

$S=\int 2\pi Tanx\sqrt{1+(Sec^2x)^2}dx$

After this step I'm lost. This is the first problem in this section involving trig so I don't think I'm seeing something clearly. I try and relate $Sec^2x$ to $1+Tan^2x$ but I don't really think that does anything.

**Krizalid** · January 20th 2010, 05:14 AM

Okay first, I misread the integral so all my latter work was wrong, but now, I get a correct integral and even better, a faster solution.

Your integral equals $\int_{0}^{\frac{\pi }{4}}{\sqrt{1+\cos ^{4}x}\cdot\frac{\sin x}{\cos ^{3}x}\,dx},$ which after putting $t=\cos x$ becomes $\int_{\frac{1}{\sqrt{2}}}^{1}{\frac{\sqrt{1+t^{4}} }{t^{3}}\,dt}=\int_{\frac{1}{\sqrt{2}}}^{1}{\frac{ dt}{t^{3}\sqrt{1+t^{4}}}}+\int_{\frac{1}{\sqrt{2}} }^{1}{\frac{t}{\sqrt{1+t^{4}}}\,dt}.$ Both of these integrals are fairly easy to solve, for example the first one is solved by putting $t\mapsto\frac1t,$ hence $\int_{\frac{1}{\sqrt{2}}}^{1}{\frac{dt}{t^{3}\sqrt {1+t^{4}}}}=\int_{1}^{\sqrt{2}}{\frac{t^{3}}{\sqrt {1+t^{4}}}\,dt}=\left. \frac{1}{2}\sqrt{1+t^{4}} \right|_{1}^{\sqrt{2}}=\frac{1}{2}\left( \sqrt{5}-\sqrt{2} \right).$ On the other hand, for the second one let's put $t\mapsto\sqrt t$ and then $\int_{\frac{1}{\sqrt{2}}}^{1}{\frac{t}{\sqrt{1+t^{ 4}}}\,dt}=\frac{1}{2}\int_{\frac{1}{2}}^{1}{\frac{ dt}{\sqrt{1+t^{2}}}}=\left. \frac{1}{2}\cdot \ln \left| t+\sqrt{1+t^{2}} \right| \right|_{\frac{1}{2}}^{1}=\frac{1}{2}\ln \frac{2\left( 1+\sqrt{2} \right)}{1+\sqrt{5}}.$

Finally, your integral equals $\frac{1}{2}\left( \sqrt{5}-\sqrt{2}+\ln \frac{2\left( 1+\sqrt{2} \right)}{1+\sqrt{5}} \right).$

**VitaX** · January 20th 2010, 02:41 PM

I'm confused on a couple things, one being that we haven't used logs or natural logs in class yet so I'm not sure if I should implement them now and how you got the first integral you listed.

The formula in the book says for surface area for revolution about the x-axis is

$S= \int_a^b 2\pi y \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$

**Krizalid** · January 20th 2010, 02:43 PM

Yes, you applied it well and I calculated the integral for you, what's the problem?

**VitaX** · January 20th 2010, 03:05 PM

Problem is I don't know how you went from my setting up the problem in my first post to the first step in your post. And on top of that you implement natural logs and I've never learned logs in calculus yet so I'm a little confused here.

**Krizalid** · January 20th 2010, 03:07 PM

Go and learn them, because I just can't believe you're covering these subjects without knowing what a logarithm is, those are the basics on pre-calculus.

**VitaX** · January 20th 2010, 03:08 PM

lol I know what a log is but we have yet to implement logs in any Calculus class I have taken and I'm in Calculus 2. What do you want me to do

Edit: I'll just go with setting up the problem and then the answer given in the book. Says its 3.84.

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