Use Green's Theorem to evaluate int_C (x^2-y^2+y cos x)dx+(x^2+y^2+sin x)dy

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2022-03-24

 

Answer & Explanation

RizerMix

RizerMix

Expert2023-04-25Added 656 answers

We can use Green's Theorem to evaluate the given line integral by computing the corresponding double integral over the region enclosed by the curve C.

Let's first write the integrand in the form required by Green's Theorem, that is, as the sum of the partial derivatives of two functions:

P(x,y)=x2-y2+ycos(x)
Q(x,y)=x2+y2+sin(x)

We have:

dPdy=-2y+cos(x)
dQdx=2x+cos(x)

Then, applying Green's Theorem, we obtain:

CP(x,y)dx+Q(x,y)dy=R(dQdx-dPdy)dA

where R is the region enclosed by C and dA is the area element.

Now we need to determine the limits of integration for the double integral. The region R is defined by:

1+x2y2
x0

This region is bounded on the left and right by the vertical lines x=0 and x=3, respectively, and on the top and bottom by the curves y=1+x2 and y=2, respectively. Therefore, we can write the double integral as:

R(dQdx-dPdy)dA=1203(2x+cos(x)+2y)dxdy

We can evaluate the inner integral with respect to x, treating y as a constant:

03(2x+cos(x)+2y)dx=(x2+sin(x)+2xy)03
=(3+sin(3)+23y)-(0+sin(0)+0)
=3+sin(3)+23y

Substituting this result into the outer integral, we get:

R(dQdx-dPdy)dA=12(3+sin(3)+43y)dy
=(3y+sin(3)y2+23y2)12
=103+sin(3)-3

Therefore, the value of the line integral is:

CP(x,y)dx+Q(x,y)dy=R(dQdx-dPdy)dA=103+sin(3)-3

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