Compute the scalar suface integral z z k z ds where k is the part of the sphere of radius 3 centered at the origin, that lies in the first octant. (hint: use the parametrization g(θ, φ) = h3 sin φ cos θ, 3 sin φ sin θ, 3 cos φi with domain of definition [0, π/2] × [0, π/2])

With the parameterization

take the normal vector to to be

which has magnitude

Then the integral is

well the 1 and -1 are always after the 0 and the 0 is inthe middle. s thats what i would guess. but i dont know what a slope of 2 is sorry

step-by-step explanation:

#15. square

step-by-step explanation:

because quadrilatural (idk how to spell it) means somthing that is not 3d or like, equal sides (or somthing like that   you know what i meanz)

hope this halps!

bai!