https://eee.uci.edu/12z/44265/home/MidExam-sample.pdf
Problem #5, I don't know what to do w/those points(plots)!, I was doing the r1(t-1)+r0(t) but it was taking forever, I am probably doing it wrong,..I need help setting this one up, Please.
Problem #5, I don't know what to do w/those points(plots)!, I was doing the r1(t-1)+r0(t) but it was taking forever, I am probably doing it wrong,..I need help setting this one up, Please.
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Problems with a bounded region is easier done using Green's Theorem
Greens Theorem states if
F(x,y)= (M(x,y), N(x,y))
The the integral of the lines of the bounded region, (it is a square bounded) is equal to
double integral of dN/dx-dM/dy with respect to area.
dN/dx-dM/dy= (2xy-1)-(1+2xy)= -2
double integral of -2..now we need to determine the interval of area. Since the hypotenuse is one and the it is a square, we can conclude the x and y component is sqrt(2)/2
integral ( integral -2,x,-sqrt(2)/2, sqrt(2)/2) ,y -sqrt(2)/2, sqrt(2)/2
do some easy double integrals
http://www.wolframalpha.com/input/?i=int…
work = -4.
I can't type 2*sqrt(2) as the integrand for some reason, so i took decimals.
Let me know if you need more help.
Greens Theorem states if
F(x,y)= (M(x,y), N(x,y))
The the integral of the lines of the bounded region, (it is a square bounded) is equal to
double integral of dN/dx-dM/dy with respect to area.
dN/dx-dM/dy= (2xy-1)-(1+2xy)= -2
double integral of -2..now we need to determine the interval of area. Since the hypotenuse is one and the it is a square, we can conclude the x and y component is sqrt(2)/2
integral ( integral -2,x,-sqrt(2)/2, sqrt(2)/2) ,y -sqrt(2)/2, sqrt(2)/2
do some easy double integrals
http://www.wolframalpha.com/input/?i=int…
work = -4.
I can't type 2*sqrt(2) as the integrand for some reason, so i took decimals.
Let me know if you need more help.