What is the Double Integral Problem for the given function?

[Gale]

Homework Statement

\int_{0}^{1}\int_{0}^{1} xy \sqrt{x^2 + y^2} dy dx

Homework Equations

The Attempt at a Solution

So I tried integration by parts, but I'm not really coming up with anything simpler. I also thought I could use a u substition, letting u= x^2+y^2, but then it was looking so messy that I thought it must be wrong. Its been a while since I've done calculus, so I'm just really unconfident in my approach.

 

[SammyS]

Homework Statement

\int_{0}^{1}\int_{0}^{1} xy \sqrt{x^2 + y^2} dy dx

Homework Equations

The Attempt at a Solution

So I tried integration by parts, but I'm not really coming up with anything simpler. I also thought I could use a u substitution, letting u= x^2+y^2, but then it was looking so messy that I thought it must be wrong. Its been a while since I've done calculus, so I'm just really not confident in my approach.

What do you get when you try that substitution?

Remember, when you integrate with respect to y, treat x as a constant.

What is \displaystyle \int_{0}^{1} xy \sqrt{x^2 + y^2} dy if you treat x as a constant?

 

[phyzguy]

You're on the right track with the u substitution. Bear in mind that when you do the first integral with respect to y, you can treat x as a constant. So if u = x^2 +y^2, then du =?

 

[Gale]

What do you get when you try that substitution?

Remember, when you integrate with respect to y, treat x as a constant.

What is \displaystyle \int_{0}^{1} xy \sqrt{x^2 + y^2} dy if you treat x as a constant?

The u substitution?

I got

u= x^2 + y^2, du= 2y dy

So I substituted back in and got:

\int_{0}^{1}\int_{0}^1 \frac{x}{2} \sqrt {u} du dx

After that I would integrate with respect to u first, getting:

\int_{0}^{1} \frac{x}{3} u^{3/2} ]_{0}^{1} dx then substituting back in
\int_{0}^{1} \frac{x}{3} (x^2 + y^2)^{3/2} ]_{0}^{1} dx and evaluating on the interval you get:
\int_{0}^{1} \frac{x}{3} ((x^2 + 1)^{3/2}-(x^2)^{3/2}) dx

And then this is where I started to lose faith in my approach...

 

[LCKurtz]

Don't give up. What's wrong with letting $u=x^2+1$ in the first term and just simplifying the second one? Also, it would save some writing if you would change the limits along with the substitution so you wouldn't have to back-substitute.

 

[SammyS]

The u substitution?

I got

u= x^2 + y^2, du= 2y dy

So I substituted back in and got:

\int_{0}^{1}\int_{0}^1 \frac{x}{2} \sqrt {u} du dx

After that I would integrate with respect to u first, getting:

\int_{0}^{1} \frac{x}{3} u^{3/2} ]_{0}^{1} dx then substituting back in
\int_{0}^{1} \frac{x}{3} (x^2 + y^2)^{3/2} ]_{0}^{1} dx and evaluating on the interval you get:
\int_{0}^{1} \frac{x}{3} ((x^2 + 1)^{3/2}-(x^2)^{3/2}) dx

And then this is where I started to lose faith in my approach...

Keep going !

Split that into two integrals.

For the first:

Do a subtitution

Let w=x²+1 .​

For the second:

Notice that (x^2)^{3/2}=x^3​

 

[Gale]

Also, it would save some writing if you would change the limits along with the substitution so you wouldn't have to back-substitute.

Okay, I'll come back to this in the morning.

Also, this may sound stupid, but how would I change the limits of the integral?

 

[phyzguy]

Okay, I'll come back to this in the morning.

Also, this may sound stupid, but how would I change the limits of the integral?

What he means is that when you started you were integrating from y=0 to y=1. When you do the first u substitution, u=x^2 + y^2, y = 0 corresponds to u=x^2, and y=1 corresponds to u=x^2+1. So you can change the limits and keep the definite integral in terms of u, so you have u^(3/2) evaluated from x^2 to (x^2+1), giving (x^2+1)^(3/2) - (x^2)^(3/2), and you don't need to back-substitute from u back to y. You get the same answer, but it is faster ans simpler.

 

[Gale]

Keep going !

Split that into two integrals.

For the first:

Do a subtitution

Let w=x²+1 .​

For the second:

Notice that (x^2)^{3/2}=x^3​

Okay, so continuing, I let w= x^2 + 1, dw= 2x dx If I change the the limits of the integral, then when x=1, w=2, when x= 0, w=1. So I get, \int_1^2 \frac{1}{6} (w)^{3/2}- \int_0^1 x^3 dx= \frac{1}{15} w^{5/2}\ \big]_1^2 \ -\ \frac{1}{4}x^4 \ \big]_0^1

Evaluated I got 0.0605. Did I make some sort of mistake?

 

[LCKurtz]

Okay, so continuing, I let w= x^2 + 1, dw= 2x dx If I change the the limits of the integral, then when x=1, w=2, when x= 0, w=1. So I get, \int_1^2 \frac{1}{6} (w)^{3/2}- \int_0^1 x^3 dx= \frac{1}{15} w^{5/2}\ \big]_1^2 \ -\ \frac{1}{4}x^4 \ \big]_0^1

Evaluated I got 0.0605. Did I make some sort of mistake?

No, but what's wrong with the exact answer?$$
\frac{16\sqrt 2 - 19}{60}$$

 

[Gale]

No, but what's wrong with the exact answer?$$
\frac{16\sqrt 2 - 19}{60}$$

Hmm... when when I simplified, my exact answer didn't look as pretty as yours and rather than trying to make into a single term, I just evaluated it on a calculator. Conveniently, it also takes up less space on my paper. haha.

Thank you for your help. I wasn't very confident when I first did this problem myself.

 

[LCKurtz]

Hmm... when when I simplified, my exact answer didn't look as pretty as yours and rather than trying to make into a single term, I just evaluated it on a calculator. Conveniently, it also takes up less space on my paper. haha.

Thank you for your help. I wasn't very confident when I first did this problem myself.

Lucky for you I'm not still teaching and you aren't in my class because I would have docked you a couple of points on an exam for not simplifying and giving an exact answer.

 

[Gale]

Lucky for you I'm not still teaching and you aren't in my class because I would have docked you a couple of points on an exam for not simplifying and giving an exact answer.

Good point! This is my first problem set for this class, so I'm not sure what sort of answers my teacher will want. I'm a big fan of exact answers, but I guess I was being lazy... I've been out of school for too many years I suppose. I'll try to simplify from now on.