Use Green's Theorem to evaluate the line integral along the given positively oriented curve.
integral of xy2 dx + 4x2y dy
C is the triangle with vertices (0, 0), (2, 2), and (2, 4)
Well, I have a joke for you before I solve the problem! Why don't scientists trust atoms? Because they make up everything!
Now, back to the problem. To evaluate the line integral using Green's Theorem, we need to find the curl of the vector field F = (xy^2, 4x^2y).
The curl of F is given by ∇ × F = (∂Q/∂x - ∂P/∂y), where P = xy^2 and Q = 4x^2y.
∂Q/∂x = 8xy, and ∂P/∂y = 2xy. Therefore, ∇ × F = 8xy - 2xy = 6xy.
Now, we need to find the area of the triangle C, which can be done by splitting the triangle into two smaller right triangles. The first right triangle has vertices (0, 0), (2, 2), and (2, 0), and the second right triangle has vertices (0, 0), (2, 4), and (2, 0).
The area of the first right triangle is (1/2) * base * height = (1/2) * 2 * 2 = 2 square units.
The area of the second right triangle is (1/2) * base * height = (1/2) * 2 * 4 = 4 square units.
Therefore, the total area of the triangle C is 2 + 4 = 6 square units.
Finally, by applying Green's Theorem, the line integral is equal to the double integral of the curl of F over the region bounded by the triangle C:
∫∫ (6xy) dA, where dA represents the area element.
Now, let me know if you need any further assistance or if you want another joke to lighten the mood!
To evaluate this line integral using Green's Theorem, we first need to determine the vector field that corresponds to the given differential form. In this case, the given differential form is xy^2 dx + 4x^2y dy.
Let F(x, y) = (P(x, y), Q(x, y)) be the vector field corresponding to this differential form, where P(x, y) = xy^2 and Q(x, y) = 4x^2y.
The next step is to find the partial derivatives of P(x, y) and Q(x, y) with respect to x and y:
∂P/∂x = y^2
∂Q/∂y = 4x^2
Now, we can apply Green's Theorem, which states that the line integral of a vector field F around a positively oriented simple closed curve C is equal to the double integral of the curl of F over the region D enclosed by C.
The curl of F is given by ∂Q/∂x - ∂P/∂y. Substituting the partial derivatives, we get:
curl(F) = ∂Q/∂x - ∂P/∂y = 4x^2 - y^2
Since C is a triangle with vertices (0, 0), (2, 2), and (2, 4), we need to find the region D enclosed by C. D will be the region bounded by the lines y = x, y = 2, and y = 4.
We can now apply Green's Theorem by evaluating the double integral of the curl of F over D:
∫∫ curl(F) dA = ∫∫ (4x^2 - y^2) dA
To evaluate this double integral, we can express it as an iterated integral. Since D is bounded by the lines y = x, y = 2, and y = 4, and x varies from 0 to 2, and y varies from x to 4, we have:
∫∫ curl(F) dA = ∫[0,2] ∫[x,4] (4x^2 - y^2) dy dx
Evaluating this iterated integral will give you the value of the line integral along the given curve C.
Oops. That is Nx - My
I'll let you make the change and redo the arithmetic.
Green's Theorem says that
∮CMdx+Ndy = ∫∫RMx+Ny dx dy
Nx = 8xy
My = 2xy
So, we end up with
∫[0,2]∫[x,2x] 10xy dy dx
= ∫[0,2] 5xy^2 [x,2x] dx
= ∫[0,2] 5x(4x^2-x^2) dx
= ∫[0,2] 15x^3 dx
= 5x^4 [0,2]
= 80