<span>Let r(x,y) = (x, y, 9 - x^2 - y^2)
So, dr/dx x dr/dy = (2x, 2y, 1)
So, integral(S) F * dS
= integral(x in [0,1], y in [0,1]) (xy, y(9 - x^2 - y^2), x(9 - x^2 - y^2)) * (2x, 2y, 1) dy dx
= integral(x in [0,1], y in [0,1]) (2x^2y + 18y^2 - 2x^2y^2 - 2y^4 + 9x - x^3 - xy^2) dy dx
= integral(x in [0,1]) (x^2 + 6 - 2x^2/3 - 2/5 + 9x - x^3 - x/3) dx
= integral(x in [0,1]) (28/5 + x^2/3 + 26x/3 - x^3) dx
= 28/5 + 40/9 - 1/4
= 1763/180 </span>
Answer:
product one; it has a higher success rate
Step-by-step explanation:
to find out which has a higher rate of success we can find out their percentage of success
we can divide their success amounts by test amounts
950 / 1000 = 95/100 = 95%
150/200 = 75/100 = 75%
product 1 has a higher success rate
Area of the shaded region = area of big square minus area of little square.
Here is the set up:
Let A_s = area of shaded region.
A_s = (2x + 2)(3x - 4) - [(x - 3)(x - 6)]
Take it from here.
X = 3/2
Y = -3
-6x-5(3-4x)=6
Y=3-4(3/2)
