Question

Find the point of the parabola z= x^2+y^2 which is closest to the point (3 -6 4)
B_ find the centroid of the first quare the area bounded by parabola y=x^2 and the line y= x2bar
Cadet ermine the centroid of the first quarter to area of the curve x=acos^3theta y=sine^3theta

Answer 1

The distance from (3,-6,4) to a point (x,y,z) on the paraboloid is

d^2 = (x-3)^2 + (y+6)^2 + (z-4)^2
= (x-3)^2 + (y+6)^2 + (x^2+y^2-4)^2

the partials of z are zero at (1,-2) and d^2 = 21

For centroids, just plug in to the formula. Should not be hard.