Change of variables + Jacobian = nice double integral

Added: 2026-05-13

[00:00:00]Let's see how we are going to evaluate this double integral here. Here r is the ellipse bounded by this equation in the first quadrant. Now have a look on the inside. 9 x^2 + 4 y^2 which is this part of the equation. Unfortunately this is not x2 + y square because otherwise we could just use the regular polar coordinates. But this right here is not so bad either because of course this right here we can look at it as 3x and then squared and this right here is the same as 2 y and then squared. So what we would like to do is put it inside 3x make that equal to r cosine theta and then for the 2y make that equal to r sin theta.

[00:00:50]So just imagine that this is like a new variable like was that's a new variable and then you still have the regular polar coordinates that kind of thing but you can just do like this as well.

[00:01:02]Now perhaps just for like a review and all that good stuff let me just do this in all details for you guys. Of course if you have done this a few times you can see some shortcuts but anyways right here divide both sides by three. X is equal to 1/3 R cosine theta. Divide both sides by two. Y it's equal to 12 R sin theta.

[00:01:25]Now have a look right here. We can square both sides. Square both sides.

[00:01:30]Square both sides. Square both sides.

[00:01:33]This right here gives us the inside 9x^2 + 4 y^2.

[00:01:39]Here that will be equal to r 2 * cosine squar theta plus r 2 * sin square theta.

[00:01:48]We can factor out the r 2 and then cosine square + sin square is just equal to one. So on the inside we will just get r 2. So that's the pretty much the polar coordinates.

[00:02:02]So now let's go ahead go to the r theta world. Here we have sign and then on the inside we have r squar and then don't forget the jacobian the usual polar coordinates we need the r but here because we have the 1/3 and also the 1/2 factors you just multiply them 1/3 * 1/2 is 1 / 6 and then you have the r.

[00:02:33]So this right here is the jacobian.

[00:02:35]Technically you apply the absolute value and then you have dr r d theta.

[00:02:41]And as I said I'm going to give you guys all the details in this video. So let's review how to do the jacobian real quick. Look at the x and y equation. We have partial xy partial r theta.

[00:02:57]This is the Jacobian notation which is defined as the determinant of partial of x with respect to r partial of x with respect to theta partial of y with respect to r partial of y with respect to theta.

[00:03:16]Now look at the x equation. Take the derivative with respect to r. That'll just give us one. So we get 1/3 cosine theta.

[00:03:25]Look at this. Take the partial with respect to theta. We get negative sin theta here. So that's - 1/3 r sin theta.

[00:03:34]Continue. Do the partial with respect to r just one. So 12 sin theta.

[00:03:41]Lastly, this right here gives us cosine and then we have the 12 r cosine theta.

[00:03:49]Okay. As you can see, when we do this times that, that's 1 / 6.

[00:03:55]And then we have cosine squared.

[00:03:59]And then what's the r? So I'll put the r right here. 1 / 6 r cosine square theta minus this times that that's negative.

[00:04:10]So it becomes positive. And then 1 / 6 again another r sin square theta.

[00:04:18]Okay. Factor out the 1 / 6 r cosine square plus sin square gives us one. So as I said this right here is the jacobian that goes right here. Very nice. Now look at the r. Have a look.

[00:04:36]This thing is exactly our r square.

[00:04:40]You can take the square root on both sides but it's just going to be one. And remember it's zero to one because we are in the first quadrant only 0 to one. And uh I'll just put that right here.

[00:04:51]And if not, if it's the whole ellipse, you can still do 0ero to one. You just have to be careful with the selection of your theta. Now, because it's in the first quadrant, it will just go from zero to pi /2. So that's all.

[00:05:07]So that will be the integral that we have to compute. Now, let's just go ahead and finish that. So we have 1 / 6 is just a constant multiple. Let's go ahead and put that all the way to the front. And uh we can separate the integrals because the inside here is just with r and then I'll just multiply with the integral with the theta. So look at the integral from 0 to pi / 2 and we have d theta because there's no theta in the function here. So we can do that and then we multiply by let's do this one in blue. the integral going from 0 to one and we have r. r is positive so we don't need the absolute value anymore and then we have sine of r² and then that is dr r.

[00:05:57]So we have 1 / 6 and then integrating this the result is just going to be pi / 2.

[00:06:04]Now for this integral I'll just do this on the side for you guys. do a u sub.

[00:06:10]Let u equ= r 2. Then we see d u is equal to 2 r d. And then d equals du over 2r.

[00:06:22]Now take this integral to the u world. r goes from zero to one. If you put zero in here, we get u is equal to zero. If you put one in here, u is equal to 1.

[00:06:33]And then we have the r sine of r² which is the u and dr r is this.

[00:06:43]As you can see r and r cancel.

[00:06:47]So 12 integrating sign we get negative cosine and then we have the u and we have to plug in zero and one. Plugging one we have - one2 and just keep it as cosine of one. This is the first term minus the second term plugging zero. So we have 12 and the negative cosine of zero. But this cosine 0 is just equal to one. So now I will write this down right here. we have - one2 cosine of one and then plus 12.

[00:07:33]That's pretty much it. And then finally, we can just factor out the 1/2 and then multiply everything.

[00:07:39]Pi on the top two * this two and then times six all together we get 24. And then this right here, we have negative cosine of one and then plus one.

[00:07:53]Yep. Just like this.

[00:07:56]And we're done.