Respuesta :
Answer:
[tex]\int_C F . dr = \pi[/tex]
[tex]C : x = cost , y = sin t, z = sin^2 t - cos^2 t , 0 \leq t \leq 2 \pi[/tex]
Step-by-step explanation:
Given that:
[tex]F(x,y,z) = x^2yi + \dfrac{1}{3}x^3j +xyk[/tex]
Here C is the curve of intersection of the hyperbolic parabolic [tex]z = y^2 - x^2[/tex] and the cylinder [tex]x^2 +y^2 =1[/tex]
Using Stokes' Theorem
[tex]\int_C F . dr =\int \int \limits_s \ curl \ F. \ds[/tex]
From above ;
S = the region under the surface [tex]z = y^2 -x^2[/tex] and above the circle [tex]x^2+y^2 =1[/tex]
Suppose, we consider [tex]f(x,y,z) =z-y^2+x^2[/tex]
therefore, S will be the level curve of f(x,y,z) = 0
Recall that:
[tex]\bigtriangledown f (x,y,z)[/tex] is always normal to the surface S at the point (x,y,z).
∴
This implies that the unit vector [tex]n = \dfrac{\bigtriangledown f}{|| \bigtriangledown ||}[/tex]
So [tex]\bigtriangledown f = <2x, -2y,1 >[/tex]
Also, [tex]|| \bigtriangledown f ||= \sqrt{4x^2+4y^2+1}[/tex]
Similarly ;
[tex]curl \ F = \begin {vmatrix} \begin{array} {ccc}{\dfrac{\partial }{\partial x} }&{\dfrac{\partial }{\partial y} }& {\dfrac{\partial }{\partial z} }\\ \\ x^2y& \dfrac{1}{3}x^3&xy \end {array} \end{vmatrix}[/tex]
[tex]curl \ F = \langle x ,-y,0 \rangle[/tex]
Then:
[tex]\int \int_s curl \ F .ds = \int \int_s curl \ F .nds[/tex]
[tex]\int \int_s curl \ F .ds = \iint_D curl \ F \dfrac{\bigtriangledown f}{ || \bigtriangledown f||} \sqrt{ (\dfrac{\partial z}{\partial x }^2) + \dfrac{\partial z}{\partial x }^2)+1 } \ dA[/tex]
[tex]\int \int_s curl \ F .ds = \iint_D \dfrac{\langle x,-y,0 \rangle * \langle 2x,-2y,1 \rangle }{\sqrt{4x^2 +4y^2 +1 }} \times \sqrt{4x^2 +4y^2 +1 }\ dA[/tex]
[tex]\int \int_s curl \ F .ds = \iint_D (2x^2 + 2y^2) \ dA[/tex]
[tex]\int \int_s curl \ F .ds = 2 \iint_D (x^2 + y^2) \ dA[/tex]
[tex]\int \int_s curl \ F .ds = 2 \int \limits ^{2 \pi} _{0} \int \limits ^1_0r^2.r \ dr \ d\theta[/tex]
converting the integral to polar coordinates
This implies that:
[tex]\int \int_s curl \ F .ds = 2 \int \limits ^{2 \pi} _{0} \int \limits ^1_0r^2.r \ dr \ d\theta[/tex]
⇒ [tex]\int_C F . dr = 2(\theta) ^{2 \pi} _{0} \begin {pmatrix} \dfrac{r^4}{4}^ \end {pmatrix}^1_0[/tex]
[tex]\int_C F . dr = 2(2 \pi) (\dfrac{1}{4})[/tex]
[tex]\int_C F . dr =(4 \pi) (\dfrac{1}{4})[/tex]
[tex]\int_C F . dr = \pi[/tex]
Therefore, the value of [tex]\int_C F . dr = \pi[/tex]
The parametric equations for the curve of intersection of the hyperbolic paraboloid can be expressed as the equations of the plane and cylinder in parametric form . i.e
[tex]z = y^2 - x^2 \ such \ that:\ x=x , y=y , z = y^2 - x^2[/tex]
[tex]x^2 +y^2 =1 \ such \ that \ : x = cos \ t , y= sin \ t, z = z, 0 \leq t \leq 2 \pi[/tex]
Set them equal now,
the Parametric equation of [tex]C : x = cost , y = sin t, z = sin^2 t - cos^2 t , 0 \leq t \leq 2 \pi[/tex]
