Evaluate the integral i = z s f ds when f(x, y, z) = z 2 + 3xy and s is the portion of the plane x + 2y + 2z = 0 above the unit disk x 2 + y 2 ≤ 1 in the xy-plane

Parameterize by

with and . The surface element is

Then the integral is

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .

yx = 36

step-by-step explanation:

cos f = opposite side/ hypotenuse

cos f = yw / yx

we know cos f = 4/9 and yw =16

4/9 = 16/yx

multiply each side by yx

4/9 yx = 16/ yx * yx

4/9 yx = 16

multiply each side by 9/4

9/4 * 4/9 yx = 16 *9/4

yx = 36