Question

Use Stokes' Theorem to evaluate

Answer 1

Answer:

$\int_C F . dr = \pi$

$C : x = cost , y = sin t, z = sin^2 t - cos^2 t , 0 \leq t \leq 2 \pi$

Step-by-step explanation:

Given that:

$F(x,y,z) = x^2yi + \dfrac{1}{3}x^3j +xyk$

Here C is the curve of intersection of the hyperbolic parabolic $z = y^2 - x^2$ and the cylinder $x^2 +y^2 =1$

Using Stokes' Theorem

$\int_C F . dr =\int \int \limits_s \ curl \ F. \ds$

From above ;

S = the region under the surface $z = y^2 -x^2$ and above the circle $x^2+y^2 =1$

Suppose, we consider $f(x,y,z) =z-y^2+x^2$

therefore, S will be the level curve of f(x,y,z) = 0

Recall that:

$\bigtriangledown f (x,y,z)$ is always normal to the surface S at the point (x,y,z).

∴

This implies that the unit vector $n = \dfrac{\bigtriangledown f}{|| \bigtriangledown ||}$

So $\bigtriangledown f =$

Also, $|| \bigtriangledown f ||= \sqrt{4x^2+4y^2+1}$

Similarly ;

$curl \ F = \begin {vmatrix} \begin{array} {ccc}{\dfrac{\partial }{\partial x} }&{\dfrac{\partial }{\partial y} }& {\dfrac{\partial }{\partial z} }\\ \\ x^2y& \dfrac{1}{3}x^3&xy \end {array} \end{vmatrix}$

$curl \ F = \langle x ,-y,0 \rangle$

Then:

$\int \int_s curl \ F .ds = \int \int_s curl \ F .nds$

$\int \int_s curl \ F .ds = \iint_D curl \ F \dfrac{\bigtriangledown f}{ || \bigtriangledown f||} \sqrt{ (\dfrac{\partial z}{\partial x }^2) + \dfrac{\partial z}{\partial x }^2)+1 } \ dA$

$\int \int_s curl \ F .ds = \iint_D \dfrac{\langle x,-y,0 \rangle * \langle 2x,-2y,1 \rangle }{\sqrt{4x^2 +4y^2 +1 }} \times \sqrt{4x^2 +4y^2 +1 }\ dA$

$\int \int_s curl \ F .ds = \iint_D (2x^2 + 2y^2) \ dA$

$\int \int_s curl \ F .ds = 2 \iint_D (x^2 + y^2) \ dA$

$\int \int_s curl \ F .ds = 2 \int \limits ^{2 \pi} _{0} \int \limits ^1_0r^2.r \ dr \ d\theta$

converting the integral to polar coordinates

This implies that:

$\int \int_s curl \ F .ds = 2 \int \limits ^{2 \pi} _{0} \int \limits ^1_0r^2.r \ dr \ d\theta$

⇒ $\int_C F . dr = 2(\theta) ^{2 \pi} _{0} \begin {pmatrix} \dfrac{r^4}{4}^ \end {pmatrix}^1_0$

$\int_C F . dr = 2(2 \pi) (\dfrac{1}{4})$

$\int_C F . dr =(4 \pi) (\dfrac{1}{4})$

$\int_C F . dr = \pi$

Therefore, the value of $\int_C F . dr = \pi$

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

$z = y^2 - x^2 \ such \ that:\ x=x , y=y , z = y^2 - x^2$

$x^2 +y^2 =1 \ such \ that \ : x = cos \ t , y= sin \ t, z = z, 0 \leq t \leq 2 \pi$

Set them equal now,

the Parametric equation of $C : x = cost , y = sin t, z = sin^2 t - cos^2 t , 0 \leq t \leq 2 \pi$