00:01
Let's have a look at the question.
00:03
So we have double integration of 2x square da.
00:08
Now we have been given the transformation x is equals to 2u and y is equals to 3v.
00:16
So let us find the jacobian which is equals to determinant of dx with respect to du differentiation and differentiation of x with respect to v.
00:29
Then we have differentiation of y with respect to u and differentiation of y with respect to v.
00:36
So this will be equals to 2 .0 and 0 3.
00:43
So here jacobian j will be equals to 6.
00:50
Now we have the ellipse as 9x square plus 4y square is equal to 30.
01:02
So let us put x is equal to 2 u and y is equals to 3v so we get 9 into 2 u whole square plus 4 into 3v whole square is equals to 36 so we get 36 u square plus 36 v square is equals to 36 so from here we can write u square plus to 1.
01:34
Now the value of u ranges from minus 1 to positive 1 and the value of v will be range from minus under root 1 minus u square to positive under root 1 minus u square.
01:56
So here we can write that double integration over the region r of 2x square d a is equals to integration from u is equal to minus 1 to positive 1 and v is equals to minus under root 1 minus u square to positive under root 1 minus u square 2 into 2 u square whole square into jacobian j d u and d v so this will be equals to 8 integration from minus 1 to positive 1 and minus under root 1 minus u square to positive 1 minus u square in integration of u square into 6 d u and dv so this will be equals to 48 integration from minus 1 to positive 1 and integration from minus under root 1 minus u square to positive under root 1 minus u square of u square dv d u.
03:09
So here we will use the polar form that is u is equal to r cos theta and v is equals to r sine theta.
03:24
So we know that u square plus v square is equal to 1...