Double Integration using u-substitution

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!

Sourabh N

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

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

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).

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.

Sourabh N

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

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.

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 shitting 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.

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?

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.

ShaneOH

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

Sourabh N

No problem :)

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?

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.

Sourabh N

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