How to do this surface integral

[jegues]

Homework Statement

See figure attached for problem statement.

Homework Equations

The Attempt at a Solution

See figure attached for my attempt so far.

I'm confused as to how to do this problem. Either plane I project the surface into (xz or yz) the integral looks pretty nasty.

Any ideas?

 

[vela]

I think it just looks worse than it really is. Try using the substitution u=4x²+1.

I should point out I'm assuming the integral you got is correct. I don't remember this stuff off the top of my head, so if there is an error, I hope someone chimes into correct us both!

 

[jegues]

I think it just looks worse than it really is. Try using the substitution u=4x²+1.

I should point out I'm assuming the integral you got is correct. I don't remember this stuff off the top of my head, so if there is an error, I hope someone chimes into correct us both!

If I apply that substitution,

\frac{1}{8} \int_{0} ^{2} \int_{1} ^{17} (4 + \frac{1-u}{4}) \sqrt{u}du dz

Simplifying,

\frac{1}{32} \int_{0} ^{2} \int_{1} ^{17} (17u^{\frac{1}{2}}-u^{\frac{3}{2}})dudz

It doesn't look like it's getting much better unless I've made a mistake.

 

[I like Serena]

I'm confused as to how to do this problem. Either plane I project the surface into (xz or yz) the integral looks pretty nasty.

Any ideas?

Your integral looks right, although I'd replace dA with dxdz.

To solve "nasty" integrals we have computers nowadays, or more specifically websites.
It does take the challenge away though .

See: http://www.wolframalpha.com/input/?i=integral+from+0+to+2%3A+x%284-x^2%29sqrt%281%2B4x^2%29+dx

Where you can find:

\int_0^2 x (4-x^2) \sqrt{1+4 x^2} dx = 1/120 (289 \sqrt 17-41) \approx 9.58815

Of course, this still needs to be integrated over z...

[EDIT]Btw, the last integral is not that difficult with e.g.:

\int x^{\frac 3 2} dx = \frac 2 5 x^{\frac 5 2} + c

[/EDIT]

 

[jegues]

Your integral looks right, although I'd replace dA with dxdz.

To solve "nasty" integrals we have computers nowadays, or more specifically websites.
It does take the challenge away though .

See: http://www.wolframalpha.com/input/?i=integral+from+0+to+2%3A+x%284-x^2%29sqrt%281%2B4x^2%29+dx

Where you can find:

\int_0^2 x (4-x^2) \sqrt{1+4 x^2} dx = 1/120 (289 \sqrt 17-41) \approx 9.58815

Of course, this still needs to be integrated over z...

[EDIT]Btw, the last integral is not that difficult with e.g.:

\int x^{\frac 3 2} dx = \frac 2 5 x^{\frac 5 2} + c

[/EDIT]

Well I hope I've made a mistake then because I am supposed to be able to do questions of this difficulty in a test situation without the use of a calculator, let alone a computer.

Can someone see where I went wrong? Or a cleaner way of doing this?

Thanks again!

 

[vela]

\frac{1}{32} \int_{0}^{2} \int_{1}^{17} (17u^{\frac{1}{2}}-u^{\frac{3}{2}})dudz

It doesn't look like it's getting much better unless I've made a mistake.

I think you're missing the obvious here. That integral is straightforward to evaluate.

What are

\int u^{1/2}\,du

and

\int u^{3/2}\,du

equal to?

 

[jegues]

I think you're missing the obvious here. That integral is straightforward to evaluate.

What are

\int u^{1/2}\,du

and

\int u^{3/2}\,du

equal to?

I know how to that do that I just don't think we would be given a question were we have ugly numbers like,

17^{\frac{5}{2}}

etc, as the answer, this question was from an old final exam and it's not like the professors to do that.

 

[LCKurtz]

I know how to that do that I just don't think we would be given a question were we have ugly numbers like,

17^{\frac{5}{2}}

etc, as the answer, this question was from an old final exam and it's not like the professors to do that.

Really? I think that number is kind of pretty. I know professors that have given problems with answers like that on exams.

 

[jegues]

Really? I think that number is kind of pretty. I know professors that have given problems with answers like that on exams.

So there's no mistakes in my original integral than? Vela said that he had skipped over that part so I'd like to make sure that what I'm doing is indeed correct.

Can you spot any errors?

 

[LCKurtz]

So there's no mistakes in my original integral than? Vela said that he had skipped over that part so I'd like to make sure that what I'm doing is indeed correct.

Can you spot any errors?

Your original xz integral is set up correctly. And the number 17^5/2 does appear in the answer, although that particular term might be simplified to 289*sqrt(17).