Question

Find ∫ C

​
F
⋅d r
where C is a circle of radius 2 in the plane x+y+z=6, centered at (1,4,1) and oriented clockwise when viewed from the origin, if F
=3y i
−3x j
​
+2(y−x) k
∫ C
​
F
⋅d r
= Solution:

Answer 1

The value of the given integral is 2
`\pi\)`.

To compute the line integral
`\(\int_C \mathbf{F} \cdot d\mathbf{r}\)` around the given closed curve C, we need to parameterize the curve and then evaluate the integral using the given vector field F.

The curve C is a circle of radius 2 centered at (1, 4, 1) lying in the plane x + y + z = 6. We can parameterize this circle using polar coordinates:

`\[x &= r\cos(\theta) + 1 \\y &= r\sin(\theta) + 4 \\z &= 6 - x - y\end{align*} \]`

where r = 2is the radius.

Now, differentiate these expressions with respect to
`\(\theta\)` to obtain
`\(\frac{d\mathbf{r}}{d\theta}\)`.

`\frac{d\mathbf{r}}{d\theta} &= \left((dx)/(d\theta), (dy)/(d\theta), (dz)/(d\theta)\right) \\&`

`= \left(-2\sin(\theta), 2\cos(\theta), -1\right)`

Now, compute
`\(\mathbf{F} \cdot \frac{d\mathbf{r}}{d\theta}\)` and integrate over the parameter interval
`\([0, 2\pi]\)` since the curve is oriented clockwise:

`\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^(2\pi) \mathbf{F} \cdot \frac{d\mathbf{r}}{d\theta} \, d\theta \]`

`\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^(2\pi) (3\sin(\theta) + 6\cos(\theta) - 2(\sin(\theta) + \cos(\theta))) \, d\theta \]`

Evaluate this integral to get the numerical value. Note that
`\(\sin\theta\)` and
`\cos\theta\)` have periodic behavior, so the integral simplifies.

The result is approximately 2
`\pi\)` or 6.283.