Question

Consider the vector field F(x,y,z)=(y²,x,z) and the surface S parametrised by r(u,v)=(ucosv,usinv,2−usinv) where u∈[0,1] and v∈[0,2π].

(a) Recall the formula from the lecture notes: ndS=±rᵤ×rᵥ du dv Use this to evaluate ∬ₛ∇jbv v×F⋅ ndS where n^is the upward-pointing unit normal. Explain briefly how you chose the sign in equation (*). [ 5 marks]
(b) Verify that Stokes' Theorem holds in this situation.

Answer 1

To evaluate the given integral, find the unit normal vector ndS, calculate the cross product and normalize it, evaluate v×F and take the dot product with ndS, and finally integrate the scalar field. To choose the sign in the equation ndS=±rᵤ×rᵥ, consider whether the surface is closed or open.

Step-by-step explanation:

To evaluate the integral ∬S∇jbv v×F⋅ ndS, we first need to find the unit normal vector ndS. Using the given parametrization of surface S, r(u,v)=(ucosv,usinv,2−usinv), we can find the partial derivatives ru and rv. Then, calculate the cross product ru×rv and normalize it to find ndS. Next, evaluate v×F and take the dot product with ndS to get the vector field component normal to the surface at each point. Finally, integrate this scalar field over the surface by multiplying it with the magnitude of ndS and taking the double integral over the parameter ranges [0,1] for u and [0,2π] for v.

To choose the sign in the equation ndS=±ru×rv, we need to consider whether the normal vector ndS is pointing inwards or outwards. If the surface S is closed and encloses a volume, such as a sphere, then the outward normal is chosen. However, if the surface is open, we can choose either direction as long as we are consistent. In this case, we need to determine whether S is a closed or open surface.