integral over c of (2x+3y)dx + (x-y)dy
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One must first create a parametrized equation for x and y using the circle C. This is easily done by using
x:= cos(t)
y:=sin(t)
noting that 0≤t≤2π
which means also that
dx:=-sin(t)dt
dy:=cos(t)dt
making appropriate substitutions, we change our integral to:
Integral from 0 to 2π of (2*cos(t)+3*sin(t))*(-sin(t)dt) + (cos(t)-sin(t))*(cos(t)dt)
=integral from 0 to 2π of {-2*sin(t)cos(t)-3sin^2(t)+cos^2(t)-sin(…
=integral from 0 to 2π of (-3sin(t)cos(t)-3sin^2(t)+cos^2(t))dt
=-3/2*cos^2(t)+3/2*sin(t)cos(t)-3/2*x+… |x=(0..2π)
=-2π
x:= cos(t)
y:=sin(t)
noting that 0≤t≤2π
which means also that
dx:=-sin(t)dt
dy:=cos(t)dt
making appropriate substitutions, we change our integral to:
Integral from 0 to 2π of (2*cos(t)+3*sin(t))*(-sin(t)dt) + (cos(t)-sin(t))*(cos(t)dt)
=integral from 0 to 2π of {-2*sin(t)cos(t)-3sin^2(t)+cos^2(t)-sin(…
=integral from 0 to 2π of (-3sin(t)cos(t)-3sin^2(t)+cos^2(t))dt
=-3/2*cos^2(t)+3/2*sin(t)cos(t)-3/2*x+… |x=(0..2π)
=-2π
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I would like to note for the record that my answer is NOT the same as pyro's. Computing antiderivatives is a needlessly ugly way to solve this problem; please don't associate that solution with my name.
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By Green's theorem, the given integral is equal to the integral over the unit disk of the function
d(x-y)/dx - d(2x+3y)/dy = 1 - 3 = -2.
Since the unit disk has area π, the answer is -2π.
If you want to do it the hard way, then parametrize the circle by x = cos t, y = sin t, for -π < t < π. Then dx = -sin t dt and dy = cos t dt. Plugging these into the given integral reduces it to the problem of
(integral from -π to π) of [-3 cos t sin t - 3 sin^2 t + cos^2 t] dt.
You can see that -3 cos t sin t is an odd function, so its integral is 0. This reduces the problem to
(integral from -π to π) of [- 3 sin^2 t + cos^2 t] dt.
We know from single-variable calculus that
(integral from -π to π) of sin^2 t dt = (integral from -π to π) of cos^2 t dt = π.
(If you don't know why this is true, the easiest way to prove it is to observe that the integrals must be equal because they both go through a full period, and that their sum must be 2π by the Pythagorean identity.) Plugging this in above, it follows that the answer is -2π.
d(x-y)/dx - d(2x+3y)/dy = 1 - 3 = -2.
Since the unit disk has area π, the answer is -2π.
If you want to do it the hard way, then parametrize the circle by x = cos t, y = sin t, for -π < t < π. Then dx = -sin t dt and dy = cos t dt. Plugging these into the given integral reduces it to the problem of
(integral from -π to π) of [-3 cos t sin t - 3 sin^2 t + cos^2 t] dt.
You can see that -3 cos t sin t is an odd function, so its integral is 0. This reduces the problem to
(integral from -π to π) of [- 3 sin^2 t + cos^2 t] dt.
We know from single-variable calculus that
(integral from -π to π) of sin^2 t dt = (integral from -π to π) of cos^2 t dt = π.
(If you don't know why this is true, the easiest way to prove it is to observe that the integrals must be equal because they both go through a full period, and that their sum must be 2π by the Pythagorean identity.) Plugging this in above, it follows that the answer is -2π.
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x²+y²=1
x=cost
y=sint
0≤t≤2π
∫ (2cost+3sint)(-sint) +(cost-sint)(cost) dt
∫ -2sintcost -3sin²t +cos²t-sintcost dt
∫ (-3/2)sin2t +1/2 +(cos2t)/2 -3/2 +(3/2)cos2t dt
∫ (-3/2)sin2t +2cos2t -1 dt
(3/4)cos2t +sin2t -t
from 2π to 0:
[(3/4)cos(4π)+sin(4π)-2π]-[(3/4)cos(0)… = -2π
x=cost
y=sint
0≤t≤2π
∫ (2cost+3sint)(-sint) +(cost-sint)(cost) dt
∫ -2sintcost -3sin²t +cos²t-sintcost dt
∫ (-3/2)sin2t +1/2 +(cos2t)/2 -3/2 +(3/2)cos2t dt
∫ (-3/2)sin2t +2cos2t -1 dt
(3/4)cos2t +sin2t -t
from 2π to 0:
[(3/4)cos(4π)+sin(4π)-2π]-[(3/4)cos(0)… = -2π