Evaluating a Surface Integral: How to Solve a Tricky Integration Problem?

[radji]

Homework Statement

It is evaluating a surface integral.

Homework Equations

∫_s∫ f(x,y,z) dS = ∫_R∫ f[x,y,g(x,y)]√(1+[g_x(x,y)]²+[g_y(x,y)]²) dA

The Attempt at a Solution

I set z=g(x) and found my partial derivatives to be g_x=√x, and g_y=0. I then inserted them back into the radical and came up with √(1+x). After integrating with respect to y (dydx) I had the final integral of 2/3 ∫ x^5/2(1+x)^1/2 dx. Instructor said to do integration by parts twice, which I've done and it still is a non integrable function. U and V keep increasing/decreasing their exponents without simplifying. I can't get wolfram alpha or symbolab to give me an answer either.

 

[haruspex]

Homework Statement

It is evaluating a surface integral.

Homework Equations

∫_s∫ f(x,y,z) dS = ∫_R∫ f[x,y,g(x,y)]√(1+[g_x(x,y)]²+[g_y(x,y)]²) dA

The Attempt at a Solution

I set z=g(x) and found my partial derivatives to be g_x=√x, and g_y=0. I then inserted them back into the radical and came up with √(1+x). After integrating with respect to y (dydx) I had the final integral of 2/3 ∫ x^5/2(1+x)^1/2 dx. Instructor said to do integration by parts twice, which I've done and it still is a non integrable function. U and V keep increasing/decreasing their exponents without simplifying. I can't get wolfram alpha or symbolab to give me an answer either.

Have you tried a trig substitution?

 

[radji]

Have you tried a trig substitution?

Yes sir, I have. But I couldn't get the √(1+x) to change to a single trig function since its 1+x.

 

[haruspex]

Yes sir, I have. But I couldn't get the √(1+x) to change to a single trig function since its 1+x.

What substitution for x will make 1+x the square of a trig function?

 

[radji]

2 cos²(u) + 1. I've substituted x for 2 cos²(u) + 1. It gets rid of the (x+1)^1/2, but I also get that identity within the 2/3x^5/2. And I can't get that expression to simplify out now.

 

[haruspex]

2 cos²(u) + 1. I've substituted x for 2 cos²(u) + 1. It gets rid of the (x+1)^1/2, but I also get that identity within the 2/3x^5/2. And I can't get that expression to simplify out now.

Clearly, it will help if x is also the square of a trig function. I can think of two substitutions that will achieve both. One uses hyperbolic trig functions, but the other is very widely used in integrals.

 

[radji]

Clearly, it will help if x is also the square of a trig function. I can think of two substitutions that will achieve both. One uses hyperbolic trig functions, but the other is very widely used in integrals.

Thank you for this. I did come up with a reasonable answer using 1+tan²u=sec²u.
It still was not the correct answer, but at least I know it how to integrate such a function now. I've attached my work so you all can let me know where else I am miscalculating. Thank you!

Spoiler

https://lh3.googleusercontent.com/NFVj3nYZrfy8V3XCCiLjnFeN_FbRbxxStVPkryH_dVVHzTPy1Iv3wNmu5HYBFDFVLhFEPxp2q9TcEog9f54qiAWBTAUJtcIpfCO8Hfm185IBQoY_hqqc2sO1Vm05_-m5-ddXtDlH6E7sPeXX6bOZfDsOwZhDqWOv-IoJn5ciGE5sDiNta86vrP9eQbA2zDrb6al1Zf5Kdu5Ne_9cDOe4l913_HGws144IT0fQwoAUb5EWdz1Slm0ARsllpj-Abtqmr3oVxxLBohohZEv3F-KcaApOjdkRpsbLRali_VEfaQDdgl1D2VDInb46D7NTCitR2ZwXOyrTOCh4-oo2OhCEMRpD8QaG_3if7U7yEYFg8MaPVzJ_Z0goZZ3laO2G70Gk_2fL5vl6z3EtkuS8sZMiBLE4-L74LlRausEZYHRZTMesm25FdGcTE43YH9FFJeRjdXI1lNzhRXyFVhBM05NT8LD7U66eHFwrdCzL9rrowQrqb91b68fjAj7MiK1o-yI3Tngw0J3j22DLcl0HjL63Rgqmk2LBbI-cysEaLhmQN54mYvnq4T41o8kl3VF2EoGuWGHImVwIN3GslgX-9i0Q7rFAklYiTo=w924-h1576-no

 

[haruspex]

Thank you for this. I did come up with a reasonable answer using 1+tan²u=sec²u.
It still was not the correct answer, but at least I know it how to integrate such a function now. I've attached my work so you all can let me know where else I am miscalculating. Thank you!

You failed to take the square root of sec².

 

[radji]

You're absolutely right. That leaves me with tan⁶u * sec³u. Wondering how that can be integrated...

 

[haruspex]

You're absolutely right. That leaves me with tan⁶u * sec³u. Wondering how that can be integrated...

With an expression like that, I look for a way of getting rid of the trig by using the standard equations sin(x).dx=-d cos(x), etc.
If you pick the right one, you should get something that you can split up into standard integrals using partial fractions, but I warn you, it looks messy.

Edit: just realized there are web pages in plenty where you can just plug in your rational function and it will decompose into partial fractions for you. That makes it quite easy.

 

[radji]

Thank you for your help. I ended up using a website to decompose it to partial fractions. It was a very messy solution.

$u=\frac 1 {288} (61\sqrt 2 - 15 sinh^{-1}(1))$