0:00
Okay.
00:02
So in this question, you're trying to integrate over some three -dimensional region.
00:07
Of course, when you're trying to integrate over a three -dimensional region, the best thing to do is try to visualize it first and then carve out that region and then strategically plan how you integrate.
00:18
So here, you have four different conditions here.
00:22
Z -graded in zero, z squared equal to x squared plus y squared, z squared less than equal to x3 times xxxx plus y squared, and then x squared plus y squared but z squared is less than equal to a squared.
00:32
Let's go over them one by one.
00:34
Z graded in zero, that just means that in your x, y, z coordinate system, you're taking basically the upper half plane, hyperplane, basically just everything with a positive z value.
00:48
The second, z squared is grader equal to x, versus y squared.
00:52
If you switch between the cartesian and polar coordinates, you see that x plus y squared is really just r squared.
01:00
Meaning, this is actually like a double -sided cone, where you have, as the radius increases, you're permitting all the z values inside the cone here.
01:15
Or you're permitting your z -absolid value has to be greater than equal to the radius that you're allotting here.
01:25
So the boundary of this is the absolute -5 -z equals of r.
01:31
Cone.
01:32
So it's like this upper and lower regions of this cone here.
01:36
Z squared is less than equal to three times x squared plus y squared.
01:39
Well, it's also a double -sided cone, but now we're talking about the in -between of the two cones, not upper and lower parts.
01:47
And the boundary is, a absolute i of z is equal to r square of three.
01:51
And finally, the x squared plus y square plus z squared is less than equal to a squared.
01:56
That's a ball.
01:58
And like all is basically the sphere and everything inside it up to radius a so if you combine all these four things together you can try and visualize well first of all we're only taking the upper co upper cones because of the z graded in zero and then the below the the lower boundary is going to be given by the x is equal to the absolute value of z equals to r and the upper boundary is going to be given by both the obfiz value of z equals r square of three and the surface of the sphere x squared plus y squared plus z squared equals a squared here so when you're taking this this like uh in between region here and this the fact that we have a bunch of cones and then the boundary being sphere a part of a sphere as well that screams to me spherical coordinates so we're going to switch to spherical coordinates here x y z going to row theta and fee um in terms of row the radius no matter which uh for for for permissible theta and phi, the radius go from 0 to a...