Question

Find the surface integral of the vector field 𝑭(𝑥, 𝑦, 𝑧) = (𝑥, 𝑦, 𝑧) over the part of the paraboloid 𝑧 = 1 − 𝑥 2 − 𝑦 2 with 𝑧 ≥ 0 and having normal pointing upwards. Hint: take 𝑥 and 𝑦 as independent parameters

Answer 1

The surface S can be represented by:

$r(x, y) = x i + y j + (1 − x^ 2 − y ^2 ) k,\ −1 ≤ x ≤ 1, −1 ≤ y ≤ 1$

$r_x = i − 2x k , r_y = j − 2y k$

$r_x × r_y=\begin{vmatrix} i & j&k \\ 1 & 0&-2x\\ 0 & 1&-2y \end{vmatrix}=2xi+2yj+k$

$\iint_SF\cdot ndS=\iint F\cdot (r_x × r_y)dA=\int^1_{-1} \int^1_{-1}(2x^2+2y^2+1 − x^ 2 − y ^2)dxdy=$

$=\int^1_{-1} \int^1_{-1}(x^2+y^2+1 )dxdy=\int^1_{-1} (2/3+2y^2+2)dy=$

$=4/3+4/3+4=20/3$