Question

Let f = ay i + bz j + cx k where a, b, and c are positive constants. let c be the triangle obtained by tracing out the path from (7, 0, 0) to (7, 0, 2) to (7, 6, 2) to (7, 0, 0). find f · dr

c.

Answer 1

The task involves evaluating the line integral of a vector field along a path to determine the work done by the field. Since the path does not traverse in the x-direction, only the y and z components of the vector field contribute to the integral. The calculation involves parameterizing each segment and summing up the individual integrals.

Step-by-step explanation:

The student's question involves finding f · dr along a path c in a vector field. The vector field is given as f = ay i + bz j + cx k, and the path c is a triangle with vertices at (7, 0, 0), (7, 0, 2), and (7, 6, 2). The dot product of a vector field with a differential displacement vector (dr) represents the work done by the field along that path.

To find the integral of f · dr along the path c, we would parameterize each segment of the path and evaluate the line integral. However, since the vector field has constants multiplied by either i, j, or k, and the path only moves parallel to the y and z axes, the components involving x do not contribute to the integral. Therefore, for each segment of the path, we will only consider the y and z components of the vector field. The line integral along the path is then the sum of the integrals over each segment of the triangle.

In this example, since the x-component of the field does not change and the path does not move in the x direction, the integral over the x components will be zero.

Calculation Steps:

1. Parameterize each segment of the path.
2. Evaluate the integral of f · dr over each segment.
3. Add the integrals from each segment to get the total work done by the field along path c.

Answer 2

You can either compute three line integrals, or use Stokes' theorem and compute one surface integral. I prefer the latter.

The curl of the given vector field is


`\\abla*\mathbf f(x,y,z)=-b\,\mathbf i-c\,\mathbf j-a\,\mathbf k`

Parameterize the triangular surface bounded by
`\mathcal C` - I'll call it
`\mathcal S` - by


`\mathbf s(u,v)=7\,\mathbf i+6v\,\mathbf j+2(u+v-uv)\,\mathbf k`

with
`0\le u\le1` and
`0\le v\le1`. By Stokes' theorem, we have


`\displaystyle\int_(\mathcal C)\mathbf f(x,y,z)\cdot\mathrm d\mathbf r=\iint_(\mathcal S)\\abla*\mathbf f(x,y,z)\cdot\mathrm d\mathbf S`

where
`\mathcal S` is positively oriented; that is, every vector normal to
`\mathcal S` is pointed in the positive
`x` direction. The surface element is given by


`\mathrm d\mathbf S=\mathbf s_u*\mathbf s_v\,\mathrm du\,\mathrm dv`

So our integral is


`\displaystyle\int_(v=0)^(v=1)\int_(u=0)^(u=1)(-b\,\mathbf i-c\,\mathbf j-a\,\mathbf k)\cdot((12v-12)\,\mathbf i)\,\mathrm du\,\mathrm dv`

`\displaystyle=12b\int_(v=0)^(v=1)(1-v)\,\mathrm dv=6b`