Question

Let Fequalsleft angle 2 xy plus z squared comma x squared plus yz comma 2 xz plus StartFraction y squared Over 2 EndFraction right angle and let C be the circle r​(t)equalsleft angle 6 sine t comma 4 sine t comma 3 cosine t right angle​, for 0less than or equalstless than or equals2pi. Evaluate ModifyingBelow Contour integral With Upper C Bold Upper F times d Bold r using any method.

Answer 1

`\oint_C F \,dr = 0`

Explanation:

In this problem, we have a vector field

`F = \left \langle 2x-z^2,x^2+yz,-2xz +(y^2)/(2) \right \rangle`.

We need to find the line integral

`\oint_C F \,dr`

where
`C` is a circle

`r(t) = \left \langle 2 \cos t, 4\sin t, 5 \cos t \right \rangle, \quad 0 \leq t \leq 2 \pi`.

As we can see, the vector filed
`F` is defined
`\forall (x,y,z) \in \mathbb{R}^3` and its component functions have continuous partial derivatives.

First, we need to find the curl of the vector filed
`F`.

`\text{curl} F = \begin{vmatrix}i & j & k\\(\partial)/(\partial x) & (\partial)/(\partial y) & (\partial)/(\partial z) \\2xy-z^2 & x^2 + yz & (y^2)/(2) - 2xz\\\end{vmatrix}`

Therefore,

`\text{curl}F = i \left( (\partial)/(\partial y) \left( (y^2)/(2) - 2xz\right) - (\partial)/(\partial z) \left( x^2+yz\right)\right) - j \left( (\partial)/(\partial x) \left( (y^2)/(2) - 2xz\right) - (\partial)/(\partial z) \left( 2xy-z^2\right)\right)\\ \phantom{12356} +k \left( (\partial)/(\partial x) \left( x^2+yz\right) - (\partial)/(\partial y) \left( 2xy-z^2\right)\right)`

Now, we can easily calculate the needed partial derivatives and we obtain

`\text{curl} F = i(y-y) - j(-2z + 2z) + k(2x-2x) = 0i + 0j+0k = \langle 0,0,0 \rangle`

So, the vector field
`F` is defined
`\forall (x,y,z) \in \mathbb{R}^3` , its component functions have continuous partial derivatives and
`\text{curl} F = 0` .Therefore, by a well-known theorem,
`F` is a conservative field.

Since
`C` is a closed path, we obtain that

`\oint_C F \,dr = 0`