00:01
Hello friends, in this question we need to evaluate the surface integral of the region f of x comma y comma z is equal to xy i cap plus y z plus z x k cap.
00:16
Now s is the part of the parabolite is it equal to 6 minus x square minus y square that lies above the sphere 0 less than or equal to x less than or equal to 1 and 0 less than or equal to y less than or equal to 1.
00:34
Now the surface x can be represented by the vector form r of x comma y is equal to x i cap plus y j cap plus for z we can replace 6 minus x square minus y square into k cap over the region of 0 less than or equal to x less than or equal to 1 and and 0 less than or equal to y less than or equal to 1.
01:04
So it follows that rx is equal to i cap minus 2x k cap and ry is equal to j cap minus 2y k cap.
01:18
Consequently we can write r x cross ry is equal to 2x i cap plus 2y j cap plus k cap.
01:31
Hence with q equal to x comma y where 0 less than or equal to x is less than or equal to 1 and 0 less than or equal to y less than or equal to 1, we obtain the surface integral over s, f dot n, ds is equal to double integral over q, f dot rx cross ry, into da.
02:07
So we can substitute the integral values like integral 0 to 1 and integral 0 to 1 we get 2x square by plus 12 y square minus 2x square y square minus 2 y power 4 plus 6x minus x cube minus x y square into dx, d -y...