Let S be the part of the sphere x^2 +y^2 +z^2 = 1 that lies above the cone z = sqrt(x^2 + y^2).Consider the following solution, where is the mistake? SLx² + y²ds.A student has evaluated this integral as follows.The unit normal to S is the vector N = xi+yj+zk. The angle y betweenN and k is Y=1=isCOS YCossphere. Therefore1> in spherical coordinates, so that the area elementAlso in spherical coordinates, vsino on thex² + y²=2πTπ/4+ y²x² + y²dS=do doCOS Y2πTsin o=do de0COS Oπ/4= 2π [ – In(cos ¢)]™*0= 2π ( − In(cos(π/4)) + In(cos 0))= 2π(− ln(√√2/2) – ln(1)) = −2π ln(√√2/2).

Gvien the surface area of the scalar function g(x,y)=x2+y2 . The surface S is part of the sphere…

A student has evaluated this integral as follows. The unit normal to S is the vector N = xi+yj+zk. The angle y between γ N and k is y = 0 in spherical coordinates, so that the area element Also in spherical coordinates, is 1 COS Y = 1 COS sphere. Therefore √ √x² + y²dS V 22. 2π = = x² + y² = sin on the x² + y² -do de 0 COS Y •2π sin = do de Cos = 2π [ – In(cos )] - = 2π ( - In(cos(7/4)) + In(cos 0)) = 2π(— In(√√2/2) - In(1)) = -2π In(√√2/2). = =-