# Thread: no idea for this vector calculus

1. ## no idea for this vector calculus

2. Originally Posted by szpengchao
I'll give a few hints here:

The (directed) line segments in each plane are given by y = -x + 1, z = -y + 1 and x = -z + 1.
The line integral is given by $\int_C z^2 - y^2 \, dx + \int_C x^2 - z^2 \, dy + \int_C y^2 - x^2 \, dz$.
Let's say that you first integrate along y = -x + 1. You can get the integral along z = -y + 1 simply by making the switch z --> y, y --> x and x --> z in the first. etc. So by symmetry .....

The line integral along y = -x + 1 should be blue sky. Use x as the parameter and note that dy = -dx and dz = 0 .....

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The statement of Stokes' Theorem should make it abundantly clear what this surface integral is equal to ........ You should of course check that the necessary conditions are satisifed.

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The surface forms part of a plane. You know three points in this plane, namely (1, 0, 0), (0, 1, 0) and (0, 0, 1), so you should be able to get the equation of the plane in the form ax + by + cz = d. As part of this process you will get the normal to the plane ......

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The surface integral should be blue sky. Using the expression for the normal given in the previous part, you just integrate over the appropriate region in the yz-plane. For example, that region can be defined by $0 \leq z \leq -y + 1$ and $0 \leq y \leq 1$ ......