00:01
So here we are given to problems.
00:04
So 10a, the first part of the problem, the first problem, we want to double integral over a region r of x plus y, da, where we have region r to be the region r to be the region that lies to the left of the y -axis between the circles.
00:22
Let's just try to see if we can sketch two circles.
00:24
So we have one circle here, and we have another circle, so we have that for example if this one we have a radius of 1 and don't have a radius of 2 but just to have an idea what we're working with so here's the x -axis and here's the y -axis and r is a region that lies to the left of the y -axis between the circles so they are referring to this region between the circles just highlight this there it's region r so we can describe this region r using our polar coordinates and by doing that, the a here is going to equal to r the r.
01:05
The theta, r here will go from 1 to 2 because of the radius of the circles that they are given here.
01:12
Theta here will go from pi 2 to 3 pi or 2.
01:17
And therefore, we're going to have the double integral to equal to the integral where data goes from pi over 2 to 3 pi or 2.
01:26
And the integral, where r goes from 1 to 2, of x plus y, where x is r, cosine theta, and y is r, sine theta.
01:36
Then we're going to multiply this by r, the r, the theta.
01:39
So we're going to have that this would equal to the integral from pi over 2 to 3, pi over 2, and then the integral with respect to r, if we can just pull out r, so we get cosine theta, plus sine theta, and we're multiplying this by r squared.
01:57
Used r to the third power over three being evaluated from 1 to 2 and then the data into this will be equal to the integral from pi over 2 to 3 pi over 2 all from so let's see what we get we plug in 2 well when we're plugging 2 into um r to the third power we hit 8 we we plug in 1 we get 1 so 8 minus 1 will be 7 so 7 over 3 comes up aside we have cosine theta for sine theta here is integral integral into integral and let's integrate this so we're going to have 7 over 3 and the integral of causing theta is sine theta and the integral of sine theta is negative causing theta the value is from pi 2 to 3 pi over 2 then this will cable 7 over 3 times so 3 pi over 2 sine of 3 power 2 is 0 minus causing a 3 pi or 2 which is negative 1 then minus 7 over 3, sine of pi over 2.
03:02
So i think it's all the way around.
03:03
So let's correct this out.
03:05
Let's go back here and say, sign of 3 pi with 2 is negative 1, and the cohesion of 3 pi or 2 is 0.
03:12
That's better.
03:13
And 7 over 3, and now sign of pi or 2, that's 1, minus causing piper 2, which is 0.
03:20
This is going to give us negative 7 over 3, minus 7 over 3, that's minus 4.
03:28
14 over 3.
03:30
That's what we get for the integral that you have there.
03:34
And it makes sense because in this region here, you can see that here we have x to be negative and y here is positive.
03:50
And in this region here, both them are negative.
03:54
Let's just double -check everything makes sense.
03:57
All right, we're going down to the next problem, which is 10b.
04:06
What do we have here? well, we want to evaluate the double integral of r of e to the power of x multiplied by y, the a, where r is the mean in the first quadrant and closed by the circle...