Respuesta :
Answer:
The result of the integral is 81π
Step-by-step explanation:
We can use Stoke's Theorem to evaluate the given integral, thus we can write first the theorem:
[tex]\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S[/tex]
Finding the curl of F.
Given [tex]F(x,y,z) = < yz, 4xz, e^{xy} >[/tex] we have:
[tex]curl \vec F =\left|\begin{array}{ccc} \hat i &\hat j&\hat k\\ \cfrac{\partial}{\partial x}& \cfrac{\partial}{\partial y}&\cfrac{\partial}{\partial z}\\yz&4xz&e^{xy}\end{array}\right|[/tex]
Working with the determinant we get
[tex]curl \vec F = \left( \cfrac{\partial}{\partial y}e^{xy}-\cfrac{\partial}{\partial z}4xz\right) \hat i -\left(\cfrac{\partial}{\partial x}e^{xy}-\cfrac{\partial}{\partial z}yz \right) \hat j + \left(\cfrac{\partial}{\partial x} 4xz-\cfrac{\partial}{\partial y}yz \right) \hat k[/tex]
Working with the partial derivatives
[tex]curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(4z-z\right) \hat k\\curl \vec F = \left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k[/tex]
Integrating using Stokes' Theorem
Now that we have the curl we can proceed integrating
[tex]\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S[/tex]
[tex]\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot \hat n dS[/tex]
where the normal to the circle is just [tex]\hat n= \hat k[/tex] since the normal is perpendicular to it, so we get
[tex]\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S \left(\left(xe^{xy}-4x\right) \hat i -\left(ye^{xy}-y\right) \hat j + \left(3z\right) \hat k\right) \cdot \hat k dS[/tex]
Only the z-component will not be 0 after that dot product we get
[tex]\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S 3z dS[/tex]
Since the circle is at z = 3 we can just write
[tex]\displaystyle \int\limits_C \vec F \cdot d\vec r = \int \int_S 3(3) dS\\\displaystyle \int\limits_C \vec F \cdot d\vec r = 9\int \int_S dS[/tex]
Thus the integral represents the area of a circle, the given circle [tex]x^2+y^2 = 9[/tex] has a radius r = 3, so its area is [tex]A = \pi r^2 = 9\pi[/tex], so we get
[tex]\displaystyle \int\limits_C \vec F \cdot d\vec r = 9(9\pi)\\\displaystyle \int\limits_C \vec F \cdot d\vec r = 81 \pi[/tex]
Thus the result of the integral is 81π
