Question

Consider the following line integral. C xy dx + x2y3 dy, c is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 4).

Answer 1

It looks like the integral is

`\displaystyle \int_C xy \, dx + x^2y^3 \, dy`

Let's close the loop by adding a line integral over the line segment joining (1, 4) to (0, 0). Then the closed loop is the triangular region

T = {(x, y) : 0 ≤ x ≤ 1 and 0 ≤ y ≤ 4x}

Since the integrand has no singularities over or on the boundary of T, we have by Green's theorem

`\displaystyle \int_C xy \, dx + x^2y^3 \, dy = \iint_T \left((\partial(x^2y^3))/(\partial x) - (\partial(xy))/(\partial y)\right) \, dx \, dy = \int_0^1 \int_0^(4x) (2xy^3 - x) \, dy \, dx`

Compute the double integral:

`\displaystyle \int_0^1 \int_0^(4x) (2xy^3 - x) \, dy \, dx = \int_0^1 \int_0^(4x) \left(128x^5 - 4x^2\right) \, dx = \frac{60}3`

From this result, we subtract the line integral over the extra line segment we added. Parameterize this path by

C' : {(1 - t, 4 - 4t) : 0 ≤ t ≤ 1}

The line integral over C' is

`\displaystyle \int_(C') xy \, dx + x^2y^3 \, dy = \int_0^1 (1-t)(4-4t)(-dt) + (1-t)^2(4-4t)^3(-4\,dt)`

`\displaystyle \int_(C') xy \, dx + x^2y^3 \, dy = \int_0^1 \left(256 t^5-1280 t^4+2560 t^3-2564 t^2+1288 t-260 \right) \, dt = -44`

so that the line integral over C alone is 60/3 - (-44) = 64.