Green's thereom

[jakeisthesnake]

Hello everyone!

So I have a problem that is bugging me well two problems actually but one is simply an integral problem.

Problem 1:

Use Green's thereom to solve the following intergral.
? (x^2 - y^2) dx + 2xy dy around the circle a^2= x^2 + y^2 where a is a constant.

Green's theorem is:
?M dx + N dy around closed curve C = ?? (dN/dx - dM/dy) dA where A is the region inclosed by curve C

what i have so far:

M=(x^2 - y^2)
dM/dy = -2y

N=2xy
dN/dx=2y
therefore: dN/dx - dM/dy= 2y - (-2y)=4y
?-a to a,?-(a^2-x^2)^(1/2) to (a^2-x^2)^(1/2), 4y dy dx

?-a to a, 2((a^2-x^2)^(1/2))^2 - 2(-(a^2-x^2)^(1/2))^2 dx
?-a to a, 2a^2 - 2x^2 - 2a^2 + 2x^2 dx
?-a to a, 0 dx
0
However the answer is supposedly 16a^2/3. Any pointers?

Problem #2
Same deal
?sin x cos y dx + (xy + sin y cos x) dy around the region bounded by y=x and y=x^(1/2)

M=sin x cos y
dM/dy = sin x sin y

N= yx + sin y cos x
dN/dx= y + cos x cos y
therefore dN/dx-dM/dy= y + cos x cos y - sin x sin y

?0 to 1,?(x)^(1/2) to x, y + cos x cos y - sin x sin y dy dx
?0 to 1, x^2/2 + cos x sin x + sin x cos x - x - cos x sin x^(1/2) + sin x cos x^(1/2) dx
?0 to 1, x^2/2 - x/2 + sin 2x + sin (x-x^(1/2))
x^3/6 x^2/4 -1/2 cos 2x + ?0 to 1, sin (x-x^(1/2)) dx

any pointers?

Thanks a bunch!

 

[galactus]

Hello everyone!

So I have a problem that is bugging me well two problems actually but one is simply an integral problem.

Problem 1:

Use Green's thereom to solve the following intergral.
? (x^2 - y^2) dx + 2xy dy around the circle a^2= x^2 + y^2 where a is a constant.

Green's theorem is:
?M dx + N dy around closed curve C = ?? (dN/dx - dM/dy) dA where A is the region inclosed by curve C

what i have so far:

M=(x^2 - y^2)
dM/dy = -2y

N=2xy
dN/dx=2y
therefore: dN/dx - dM/dy= 2y - (-2y)=4y
?-a to a,?-(a^2-x^2)^(1/2) to (a^2-x^2)^(1/2), 4y dy dx

?-a to a, 2((a^2-x^2)^(1/2))^2 - 2(-(a^2-x^2)^(1/2))^2 dx
?-a to a, 2a^2 - 2x^2 - 2a^2 + 2x^2 dx
?-a to a, 0 dx
0
However the answer is supposedly 16a^2/3. Any pointers?

You are correct. It is 0.

\(\displaystyle \int_{C}(x^{2}-y^{2})dx+2xydy=\int_{R}\int (2x+2y)dA=\int_{-a}^{a}\int_{-\sqrt{a^{2}-x^{2}}}^{\sqrt{a^{2}-x^{2}}}4ydydx=4\int_{-a}^{a}0dx=0\)

You could also do this with polar. It still give the same solution.......0

Problem #2
Same deal
?sin x cos y dx + (xy + sin y cos x) dy around the region bounded by y=x and y=x^(1/2)

M=sin x cos y
dM/dy = sin x sin y

N= yx + sin y cos x
dN/dx= y + cos x cos y
therefore dN/dx-dM/dy= y + cos x cos y - sin x sin y

?0 to 1,?(x)^(1/2) to x, y + cos x cos y - sin x sin y dy dx
?0 to 1, x^2/2 + cos x sin x + sin x cos x - x - cos x sin x^(1/2) + sin x cos x^(1/2) dx
?0 to 1, x^2/2 - x/2 + sin 2x + sin (x-x^(1/2))
x^3/6 x^2/4 -1/2 cos 2x + ?0 to 1, sin (x-x^(1/2)) dx

any pointers?

Thanks a bunch!

\(\displaystyle \int_{C}sin(x)cos(y)dx+(xy+cos(x)sin(y))dy=\int_{R}{\int\left[(y-sin(x)sin(y))-(-sin(x)sin(y))\right]dA\)

\(\displaystyle =\int_{0}^{1}\int_{x}^{\sqrt{x}}ydydx\)

Finish up?.