00:01
In this question, we are given a paraboloid, z is equals to 8 plus 2x squared plus 2y square and we are given z as 14 which lies in first octant.
00:15
So first let us find the region.
00:22
So let's take z as equals to 8 plus 2 times of x squared plus y square.
00:27
So x squared plus y square can be returned as r squared.
00:32
So, z is equals to 8 plus 2r squared.
00:36
So now let's substitute the value of z.
00:38
So it is going to be 14 is equals to 8 plus 2r squared.
00:44
So simplifying this, we'll be getting r as 6 is equals to 2r squared.
00:53
So r is going to be root 3.
00:55
So the region of r is going to be 0 less than or equals to r less than or equals to 3.
01:00
So now next we have to find the region of theta.
01:07
So since it is in first octant, it is going to be 0 less than or equals to teta less than or equals to pi over 2.
01:15
So now let us write this in double integral region and let us write 8 plus 2x squared plus 2y square subtracted by z.
01:36
So we'll be having d -teta and dr.
01:39
So let's substitute the values.
01:42
So it's going to be 0 to root 3 and 0 to pi over 2 and 8 plus 2r squared subtracted by z is going to be 14 for us.
01:55
So d -teta and dr.
01:58
So proceeding further, we'll be having integration of 0 to root 3 and integration of 0 pi over 2...