Homework Help: Double Integration using u-substitution

1. Jul 25, 2012

ShaneOH

1. The problem statement, all variables and given/known data
Hey guys, so we're going over multiple integration in Calc III, and I'm having trouble with the more complex problems.

∬ √(x + 4y) dxdy, where R = [0, 1] x [2, 3]

So it's 0 to 1 on the outside integral, 2 to 3 on the inside integral, of sqrt(x+4y) dxdy.

2. Relevant equations

Iterated Integrals

u-substitution

3. The attempt at a solution

I solved the previous double integration problem, but using u-substitution in two variables is throwing me off.

I would assume that, u = x + 4y
du = 1 dx (since we're integrating with respect to x first, and holding y as a constant, so x becomes 1 and 4y drops out). So du = dx

= ∬ √(u) du
= ∫(0 to 1) [(2u^(3/2))/3] (2 to 3) dy
= ∫(0 to 1) (((2(3 + 4y)^(3/2))/3) - (2(2 + 4y)^(3/2))/3)) dy

And this is where I stopped since this seems way overly complicated, and I feel like I did something wrong. My main problem:

Change of variables: If I want to change the integral variables, i.e. 2 to 3, I know in the single variable case that you use the u equation. But here, u = x + 4y, it has two variables. I only have 2 and 3 so do I just plug that into the x and completely ignore anything attached to a y? Pretend the y isn't there? It doesn't make much sense to me.

u-substitution in general, with multiple integration: I'm not sure if I'm going through the process after that correctly, either.

If I could figure this out, and know the correct way to do these kinds of problems in general, I could do much more, but for now I'm stuck on all of these problems since they're mostly similar.

Thank you very much for any help or input!

2. Jul 25, 2012

Sourabh N

Correct me if this is wrong, the point where you stopped is
$\int^{0}_{1}$ [$\frac{(3 + 4y)^{3/2}}{3/2}$ - $\frac{(2 + 4y)^{3/2}}{3/2}$] dy

You are on the right track. Use the same technique as the one used for x-integration.

3. Jul 25, 2012

daveb

Yes, you're being way too complicated. If you first integrate wrt x, just treat the 4y as some constant, so you're integrating sqrt(X+C).

4. Jul 25, 2012

ShaneOH

@Sourabh. Sure thing, it just looked... wrong, but I'll take a crack at continuing.

@daveb: I was under the impression I needed u-substitution for more than one value under a square root? Or does it just become (2(x+C)^(3/2))/3)?

Also: Could someone just shed some light on my change of variables question? That would clear things up a good deal.

5. Jul 25, 2012

Sourabh N

It'd be better if you rephrase your "Change of variables" question, making it more specific. :)

6. Jul 25, 2012

ShaneOH

Sure, so I'm confused on:

Single variable substitution: Let's say you have some arbitrary integration problem from 2 to 4, of √(x^2 + x).

Here, u = x^2 + x
du = 2x + 1 dx
dx = du/2x + 1

And so, instead of 2 to 4, the bounds become:

(2)^2 + (2) = 6

(4)^2 + 4 = 20

So the problem becomes: integrate, from 6 to 20, (sqrt(u)) du/2x+1, yes?

So I understand changing the bounds in those cases, but for example, in this problem, where u = x + 4y - I don't know what to do with the y. If I follow the same method, with bounds 2 to 3, the bounds would become:

2 + 4y to 3 + 4y

And I'm not sure how to handle the y. I don't know how to change the bounds using u-sub with multiple variables.

7. Jul 25, 2012

Sourabh N

Yes! (of course, after getting rid of the x in the denominator.)

That is correct. Imagine integrals as beasts, eating the integrand and garbageting the integrate. A dx integral cannot digest y, thus leaves it as it is. For this reason, you can treat y as a constant, like 4 or 5 or 42.

8. Jul 25, 2012

ShaneOH

So (last question!) what you're saying is, if this problem would have been such that u = 2x + 4y, then the bounds would have been from 4 to 6?

I literally just completely drop the y? Ignore it?

Say, u = 2x + xy + y, and normal bounds are from 2 to 4

Then, change of bounds: u = 2(2) + 2 and 2(4) + 4
Making it 6 to 12? Again just pretending the y isn't there even though it is attached to the x as well?

9. Jul 25, 2012

Sourabh N

Nope, don't ignore it, but "leave it as it is".
Some examples:

1. u = 42x + y; x goes from 0 to 1 => u goes from y to 42 + y.
2. u = 42x + 5; x goes from 0 to 1 => u goes from 5 to 47.
2. u = x + xy + y2; x goes from 4 to 6 => u goes from 4 + 4y + y2 to 6 + 6y + y2.

10. Jul 25, 2012

ShaneOH

So they become variable bounds. Got it. Thanks so much for the help and patience Sourabh!

11. Jul 25, 2012

Sourabh N

No problem :)

12. Jul 25, 2012

ShaneOH

Hey guys, I'm sorry to bring the same problem up again, but my answer is lacking two entire terms and I'm not sure what I'm omitting.

Problem:

∬ √(x + 4y) dxdy, where R = [0, 1] x [2, 3]

Solution attempt:

I would assume that, u = x + 4y
du = 1 dx (since we're integrating with respect to x first, and holding y as a constant, so x becomes 1 and 4y drops out). So du = dx

= ∬ √(u) du
= ∫(0 to 1) [(2u^(3/2))/3] (2 to 3) dy
= ∫(0 to 1) (((2(3 + 4y)^(3/2))/3) - (2(2 + 4y)^(3/2))/3)) dy

Moving on...

u = 3 + 4y, du = 4 dy, dy = du/4

Bounds (0 to 1) become 3 to 7:

= ∫(3 to 7) (2u^(3/2))/3) - (2u^(3/2))/3) du/4

= 1/4[(4u^(5/2)/15) - (4u^(5/2)/15)] from 3 to 7
((1/4) cancels out with both terms on inside so: )
= (7^(5/3)/15) - (3^(5/2)/15)

= 1/15(7^(5/2) - 3^(5/2))

And done. However, the back of the book disagrees with me, stating the answer as:

1/15(7^(5/2) - 6^(5/2) - 3^(5/2) + 2^(5/2))

And I looked back through my work and I just don't see where that second and fourth term are coming from. Is there a step or two I'm completely missing?

13. Jul 26, 2012

daveb

If you're integrating [0,1]x[2,3], your bounds for x range from 0 to 2 and y ranges from 1to 3. You have the limits of integration wrong.

14. Jul 29, 2012

Sourabh N

Look carefully at the boldface terms. Do you see what went wrong?