# Math Help - Surface integral with exponentials

1. ## Surface integral with exponentials

Hi, I've been looking at surface integrals, and I'm having a problem with all the exponential functions I come across. In a single variable case:

$\int_{a}^b e^{2x+1} dx$

I would just take this as:

$\int_{a}^b e^{2x+1} dx = \int_{a}^b e^{u} du = [\frac{e^u}{f'(u)}]$

and then take the integral within the new limits given by du = 2dx.

An example of a question with exponentials in a surface integral that I'm stuck on is, evaluate:

$\int_{0}^2 \int_{x}^{2x} e^{x+y} dy dx$

How do I go about such a thing? I know how to do some basic surface integrals, with indicies/trig functions etc, over rectangular and non-rectangular domains BTW, but this exponential stuff has got me beat. Thanks

P.S. how can I get the brackets properly round the fraction with Latex? Thanks

2. Originally Posted by Greengoblin
Hi, I've been looking at surface integrals, and I'm having a problem with all the exponential functions I come across. In a single variable case:

$\int_{a}^b e^{2x+1} dx$

I would just take this as:

$\int_{a}^b e^{2x+1} dx = \int_{a}^b e^{u} du = [\frac{e^u}{f'(u)}]$

and then take the integral within the new limits given by du = 2dx.

An example of a question with exponentials in a surface integral that I'm stuck on is, evaluate:

$\int_{0}^2 \int_{x}^{2x} e^{x+y} dy dx$

How do I go about such a thing? I know how to do some basic surface integrals, with indicies/trig functions etc, over rectangular and non-rectangular domains BTW, but this exponential stuff has got me beat. Thanks

P.S. how can I get the brackets properly round the fraction with Latex? Thanks
you evaluate it the same way. when you're dealing with dy, treat x as a constant, when dealing with dx treat y as a constant. that is, when doing $\int e^{x + y}~dx$, it might as well be $\int e^{x + 1}~dx$, you'd treat the y as any other constant, here i used 1, but it could be anything in your head. the same rules apply

can you continue?

3. thanks, i'll try:

$u=(x+y), \frac{du}{dy} = 1, du=dy$

$\int_{0}^2 \int_{x}^{2x} e^{x+y} dy dx = \int_{0}^2 [\frac{e^{u}}{du/dy}]_{x}^{2x} dx = \int_{0}^2 [\frac{e^{u}}{1}]_{x}^{2x} dx =\int_{0}^2 [e^{x+y}]_{x}^{2x} dx$

$= \int_{0}^2 [e^{3x} - e{2x}]dx = [\frac{e^{3x}}{3} - \frac{e^{2x}}{2}]_{0}^2 = \frac{e^6}{3} - \frac{e^4}{2} - \frac{1}{3} + \frac{1}{2}$

...and then calculate. d'ya reckon thats ok? Thanks

Sorry about the brackets I don't know how to get them around the fraction properly - for everything I try it has a latex error. Also I spose alot of the 'd's should be the wigly partial derivative sign, but I don't know how to get that either.

4. Originally Posted by Greengoblin
thanks, i'll try:

$u=(x+y), \frac{du}{dy} = 1, du=dy$

$\int_{0}^2 \int_{x}^{2x} e^{x+y} dy dx = \int_{0}^2 [\frac{e^{u}}{du/dy}]_{x}^{2x} dx = \int_{0}^2 [\frac{e^{u}}{1}]_{x}^{2x} dx =\int_{0}^2 [e^{x+y}]_{x}^{2x} dx$

$= \int_{0}^2 [e^{3x} - e{2x}]dx = [\frac{e^{3x}}{3} - \frac{e^{2x}}{2}]_{0}^2 = \frac{e^6}{3} - \frac{e^4}{2} - \frac{1}{3} + \frac{1}{2}$

...and then calculate. d'ya reckon thats ok? Thanks
you got the idea. i just think you'd want to simplify more. maybe get everything in one fraction or so

Sorry about the brackets I don't know how to get them around the fraction properly - for everything I try it has a latex error.
use \left[ \right]

this works in general to wrap brackets around anything. so you could do \left( \right) or \left \{ \right \} etc

Also I spose alot of the 'd's should be the wigly partial derivative sign, but I don't know how to get that either.
type \partial to get $\partial$

5. Thanks alot, I spose I could get everthing as a fraction over 6, but it was getting hard to decipher the latex I've already written . I've got a couple more of these questions too, but I'll have another go at them, and I'll post my working back here later. Cheers!

6. Originally Posted by Greengoblin
Thanks alot, I spose I could get everthing as a fraction over 6, but it was getting hard to decipher the latex I've already written . I've got a couple more of these questions too, but I'll have another go at them, and I'll post my working back here later. Cheers!
ok, that's good. good luck.

(you could write it out on paper first and then translate to LaTex)