00:01
Hi, here for the given question we are given that here we have t of u ,v equals to u ,v ,uv and here we are given that u square plus v square is less than or equal to 1.
00:14
Now here we need to calculate the value of the surface area using the parametrization of the surface.
00:19
So s is equal to double integration over d multiplied with here integral is norm of cross product of tu and tv here du ,dv.
00:29
Now here first of all we need to find tu.
00:32
Tu is nothing but partial derivative of this t with respect to u.
00:36
So we have 1 ,0 ,v.
00:38
Now similarly tv equals to 0 ,1 ,u.
00:42
So cross product of tu and tv can be written as here we have i cap ,j cap ,k cap and here we have 1 ,0 ,v ,0 ,1 ,u.
00:52
So this can be written as now calculating the cross product we have minus v ,minus u and 1.
00:58
So this is the value.
00:59
Now further in the next step of the question we need to calculate the value of the norm.
01:04
So norm of tu cross tv can be calculated as under root of v square plus u square plus 1.
01:13
Now we are already given that this is the value and here the limit of the integration will be for u and v being the part of the circle s can be written as integration over 0 to 2 pi integration over 0 to 1 under root of v square plus u square plus 1 and here we have du ,dv...