Surface area: y=e^x 0<=x<=1, around x-axis
- Thread starter cheffy
- Start date
Yes, your set up looks good.
You could try Integration by parts.
Let \(\displaystyle \L\\u=\sqrt{1+e^{2x}}, \;\ dv=e^{x}dx, \;\ du=\frac{e^{2x}}{\sqrt{1+e^{2x}}}, \;\ v=e^{x}dx\)
Personally, I'd use tech to do it. But this ought to give you good integration practice.
You could try Integration by parts.
Let \(\displaystyle \L\\u=\sqrt{1+e^{2x}}, \;\ dv=e^{x}dx, \;\ du=\frac{e^{2x}}{\sqrt{1+e^{2x}}}, \;\ v=e^{x}dx\)
Personally, I'd use tech to do it. But this ought to give you good integration practice.
Yes, it is rather messy, but not 'undoable'.
Using parts you get:
\(\displaystyle \L\\e^{x}\sqrt{1+e^{2x}}-\int{e^{x}\cdot\frac{e^{2x}}{\sqrt{1+e^{2x}}}\)
For the integral part, you can use substitution. u-subbing is mostly used.
For this you could let \(\displaystyle \L\\u=\sqrt{1+e^{2x}}, \;\ du=\frac{e^{2x}}{\sqrt{1+e^{2x}}}dx, \;\ \sqrt{u^{2}-1}=e^{x}\)
This gives:
\(\displaystyle \L\\\int\sqrt{u^{2}-1}du\)
For this, you could use trig sub,
Or some other substitution. Try working through this. It's good practice and maybe you'll get a sense of accomplishment when you're done.
Using parts you get:
\(\displaystyle \L\\e^{x}\sqrt{1+e^{2x}}-\int{e^{x}\cdot\frac{e^{2x}}{\sqrt{1+e^{2x}}}\)
For the integral part, you can use substitution. u-subbing is mostly used.
For this you could let \(\displaystyle \L\\u=\sqrt{1+e^{2x}}, \;\ du=\frac{e^{2x}}{\sqrt{1+e^{2x}}}dx, \;\ \sqrt{u^{2}-1}=e^{x}\)
This gives:
\(\displaystyle \L\\\int\sqrt{u^{2}-1}du\)
For this, you could use trig sub,
Or some other substitution. Try working through this. It's good practice and maybe you'll get a sense of accomplishment when you're done.
Hello, cheffy!
Let: \(\displaystyle \L\,u\,=\,e^x\;\;\Rightarrow\;\;x\,=\,\ln u\;\;\Rightarrow\;\;dx\,=\,\frac{du}{u}\)
Substitute: \(\displaystyle \L\:2\pi\int u\sqrt{1\,+\,u^2}\,\frac{du}{u}\;=\;2\pi\int\sqrt{1\,+\,u^2}\,du\)
Let: \(\displaystyle \L\,u\,=\,\tan\theta\;\;\Rightarrow\;\;du\,=\,\sec^2\theta\,d\theta\)
Substitute: \(\displaystyle \L\:2\pi\int\sec\theta(\sec^2\theta\,d\theta) \;=\;2\pi\int\sec^3\theta\,d\theta\)
This will require Integration By Parts, unless you already know the formula:
. . . . \(\displaystyle \L\frac{1}{2}\left[\sec\theta\tan\theta \,+\,\ln\left|\sec\theta\,+\,\tan\theta\right|\right]\)
Then back-substitute and evaluate . . .
Let: \(\displaystyle \L\,u\,=\,e^x\;\;\Rightarrow\;\;x\,=\,\ln u\;\;\Rightarrow\;\;dx\,=\,\frac{du}{u}\)
Substitute: \(\displaystyle \L\:2\pi\int u\sqrt{1\,+\,u^2}\,\frac{du}{u}\;=\;2\pi\int\sqrt{1\,+\,u^2}\,du\)
Let: \(\displaystyle \L\,u\,=\,\tan\theta\;\;\Rightarrow\;\;du\,=\,\sec^2\theta\,d\theta\)
Substitute: \(\displaystyle \L\:2\pi\int\sec\theta(\sec^2\theta\,d\theta) \;=\;2\pi\int\sec^3\theta\,d\theta\)
This will require Integration By Parts, unless you already know the formula:
. . . . \(\displaystyle \L\frac{1}{2}\left[\sec\theta\tan\theta \,+\,\ln\left|\sec\theta\,+\,\tan\theta\right|\right]\)
Then back-substitute and evaluate . . .