Evaluate the following integral using any Vector Calculus method

Question

The integral with bounds C (xyzi+y⁴j+(2y+z⁸)k) dot dr, where C is the intersection of the unit cube and the plane z=x/10+y/20+1/30;

Specific which method you used.

Answer 1

Let

F = <xyz, y^4, 2y + z^8>.

Since C is the closed curve formed by the intersection of the plane with the unit cube, Stokes’ Theorem is the natural method:

∮C F · dr = ∬S (curl F) · n dS.

First compute the curl:

curl F = <2, xy, -xz>.

The surface is the plane

z = x/10 + y/20 + 1/30.

Parameterize it by

r(x,y) = <x, y, x/10 + y/20 + 1/30>,

where 0 <= x <= 1 and 0 <= y <= 1.

For the upward orientation,

r_x × r_y = <-1/10, -1/20, 1>.

Therefore,

(curl F) · (r_x × r_y)

= <2, xy, -xz> · <-1/10, -1/20, 1>

= -1/5 - xy/20 - xz.

Substitute

z = x/10 + y/20 + 1/30:

= -1/5 - x^2/10 - xy/10 - x/30.

So

∮C F · dr

= integral from 0 to 1 integral from 0 to 1

[-1/5 - x^2/10 - xy/10 - x/30] dy dx.

Evaluating gives

∮C F · dr = -11/40.

This uses Stokes’ Theorem with the upward normal, so C is oriented counterclockwise when viewed from above. If C is traversed in the opposite direction, the answer changes sign to 11/40.

Final answer

Using Stokes’ Theorem: \(-11/40\) for the standard positive/upward orientation.