Gauss-Green Cardioid problem not coming out right

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Homework Statement


Here is the region R plotted in the xy plane by the functions
X(t)=cos(t)(1-cos(t))
Y(t)=sin(t)(1-cos(t))

go with
f(x,y)=3+y-x and g(x,y)=3-x

calculate ∫∫_R f(x,y) dxdy and ∫∫_R g(x,y) dxdy

Homework Equations


The Attempt at a Solution


I know I need to use the Gauss-Green formula so

X(t)=cos(t)(1-cos(t)
X'(t)=-sin(t)(-cos(t)+1)+cos(t)sin(t)

Y(t)=sin(t)(1-cos(t)
Y'(t)=cos(t)(-cos(t)+1)+sin(t)^2

a=0
b=2∏

f(x,y)=3+y-x

m(x,y)=0
n(x,y)=∫f(s,y)ds=-.5x^2+3x+xy

∫(m(X(t),Y(t))X'(t)+n(X(t),Y(t))Y'(t)) dt from a to b

Now I know this is a cardioid with a area of 3∏/2 but when I put everything in I get 18.06. Is my a and b wrong because I swear I have done everything else right
 
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delta59 said:

Homework Statement


Here is the region R plotted in the xy plane by the functions
X(t)=cos(t)(1-cos(t))
Y(t)=sin(t)(1-cos(t))

go with
f(x,y)=3+y-x and g(x,y)=3-x

calculate ∫∫_R f(x,y) dxdy and ∫∫_R g(x,y) dxdy


Homework Equations





The Attempt at a Solution


I know I need to use the Gauss-Green formula so

X(t)=cos(t)(1-cos(t)
X'(t)=-sin(t)(-cos(t)+1)+cos(t)sin(t)

Y(t)=sin(t)(1-cos(t)
Y'(t)=cos(t)(-cos(t)+1)+sin(t)^2

a=0
b=2∏

f(x,y)=3+y-x

m(x,y)=0
n(x,y)=∫f(s,y)ds=-.5x^2+3x+xy

∫(m(X(t),Y(t))X'(t)+n(X(t),Y(t))Y'(t)) dt from a to b

Now I know this is a cardioid with a area of 3∏/2 but when I put everything in I get 18.06. Is my a and b wrong because I swear I have done everything else right
The integral won't come out to be equal to the area of the cardioid. Why do you think it does? Your answer of 18.06 is correct.
 
lol wow I feel stupid, its measuring a 3d space formed by the cardiod and the plane of f(x,y) and g(x,y). I get it now
 
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