<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>
36÷6 is 6 to show your work for it maybe draw it out and better show the equation like when you solve a bigger problem.
Answer:
1. No
2. No
3. No
4. Yes
5. Yes
Step-by-step explanation:
1a) f(x) = I x+2 I. This is a piece-wise graph ( V form)
x = 0 →f(x) =2 (intercept y-axis)
x = -2→f(x) = 0 (intercept x-axis)
x = -3→f(x) = 1 (don't forget this is in absolute numbers)
x = -4→f(x) = 2 (don't forget this is in absolute numbers)
Now you can graph the V graph
1b) Translation: x to shift (-3) units and y remains the same, then
f(x-3) = I x - 3 + 2 I = I x-1 I
the V graph will shift one unit to the right, keeping the same y. Proof:
f(x) = I x-1 I . Intercept x-axis when I x-1 I = 0, so x= 1
If I could get more context then that would help. Because an exponent being an odd number doesn’t do anything, so it could be either a negative or a positive
