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Posted by julia on Monday, December 10, 2007 at 9:15pm.
Can someone show me step by step how to work this problem out please.
Let C be the triangle with vertices (0,0), (1,0), and (1,2). Find the following line integral by two methods, directly and using greens theorem.
2xy^2dx + yx^2dy
Thanks!!!
For Further Reading
* Greens Theorem - Damon, Tuesday, December 11, 2007 at 6:35pm
directly
first from (0,0) to (1,0)
y = 0
x goes from 0 to 1
both terms are zero, so no contribution
second from (1,0) to (1,2)
Here dx = 0, x=1 and y goes from 0 to 2
so from 0 to 2 of y dy
= 2^2/2 = 2 total so far
now from (1,2) to(0,0)
dy = 2dx
y = 2x
so the integrand is
2 x (2x)^2 dx + 2x^3(2dx)
= 8x^3 dx + 4x^3 dx
= 12x^3 dx
from x = 1 to x = 0
3(0)^4 -3 (1^4) = -3
2 - 3 = -1 finally
NOW, using Green
closed path integral of Ldx+Mdy = integral over surface inside of
(dM/dx -dL/dy)
here L = 2xy^2
so dL/dy = 4xy
here M = y x^2
so dM/dx =2xy
so
dM/dx - dL/dy = integrand = -2xy
integral of
-2xy dy dx from y = 0 to y = 2x
then from x = 0 to 1
-2 *int x dx *int y dy
-2 * int x dx *[ y^2/2 from y = 0 to y = 2x]
-2 int x dx *2x^2
-4 int x^3 from 0 to 1
-4 (1^4/4) = -1
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To work out the given line integral, there are two methods - directly and using Green's theorem. Let's go through both methods step by step.
Direct Method:
1. First, divide the triangle C into three segments. The first segment is from (0,0) to (1,0), the second segment is from (1,0) to (1,2), and the third segment is from (1,2) to (0,0).
2. For the first segment, y = 0, and x goes from 0 to 1. Plug these values into the integrand, which is 2xy^2dx + yx^2dy. Both terms in the integrand become zero, so there is no contribution.
3. For the second segment, dx = 0, x = 1, and y goes from 0 to 2. Substitute these values into the integrand and integrate with respect to y. You will get the integral of ydy from 0 to 2. This equals 2^2/2 = 2.
4. For the third segment, dy = 2dx and y = 2x. Substitute these values into the integrand and simplify. You will have the integrand as 8x^3 dx + 4x^3 dx, which simplifies to 12x^3 dx.
5. Integrate the integrand from x = 1 to x = 0. This will give you 3(0)^4 - 3(1^4) = -3.
6. Add up the contributions from each segment: 0 + 2 + (-3) = -1.
Using Green's Theorem:
1. Apply Green's theorem, which states that the line integral around a closed path is equal to the double integral of the curl of the vector field over the area enclosed by the path.
2. The integrand is 2xy^2dx + yx^2dy, so the vector field components are L = 2xy^2 and M = yx^2.
3. Calculate the partial derivatives: dL/dy = 4xy and dM/dx = 2xy.
4. Subtract dL/dy from dM/dx to get the integrand, which is -2xy.
5. Integrate the integrand over the region enclosed by the path. The integration limits are y = 0 to y = 2x and x = 0 to x = 1.
6. Simplify the integral to get -4 int x^3 from 0 to 1.
7. Evaluate the integral, which gives you -4(1^4/4) = -1.
So, the final answer for the line integral is -1 in both methods.