Let s be a sphere x ^ 2 + y ^ 2 + Z ^ 2 = 1, and find the value of surface integral ∫ (x + y + Z + 1) ds. The answer is 4 Π

Let s be a sphere x ^ 2 + y ^ 2 + Z ^ 2 = 1, and find the value of surface integral ∫ (x + y + Z + 1) ds. The answer is 4 Π

According to the symmetry of sphere, the integral of odd function about X, y, Z is 0
So ∫ XDS = ∫ YDS = ∫ ZDS = 0
therefore
Original integral = ∫ (x + y + Z + 1) ds = ∫ ∫ DS = surface area of sphere = 4 π