00:01
We are given with z is equals to 11 minus x square minus y square and it is given that it is bounded below by z is equals to 2.
00:14
We have to find area for this form.
00:18
So the formula is a area is equals to double integration over the domain d into integration of 1 plus in root of 1 plus derivative of z with respect to x it's a square plus derivative of z with respect to y it's a square and here is da.
00:42
Now if we derivative z with respect to x from here we get 1 plus z derivative with respect it become minus 2x and it's a square it is 4x square.
00:55
Similarly derivative of z with respect to y it become minus 2y and it's a square become 4y square that is da.
01:06
We can take out common 4 and left with 4x square plus y square with da.
01:18
Now we will use this condition that is given z is equals to 2.
01:22
If we put z equals to 2 here we get 2 is equals to 11 minus x square minus y square from here x square plus y square is equals to 9.
01:37
9 can be written as 3 square that mean r is 2 here r is 2.
01:42
Now if we change the coordinate is x is equals to r cos theta y is equals to r sin theta r sin theta and limits of r is from 0 to 3 0 to r is 3 here 0 to 3 and theta is moving from 0 to 2 pi.
02:13
We are changing our this term terms into polar coordinates and therefore da is also changed.
02:20
Da will be converted into r dr d theta.
02:25
These are the transformation when we convert into another system.
02:29
So it's a 1 plus 4 x square plus y square x is r cos theta y is r sin theta.
02:36
If we square it and add it we get only r square and da is r dr d theta...