Use Stokes' Theorem to evaluate
C
F · dr
where C is oriented counterclockwise as viewed from above.
F(x, y, z) = xyi + 5zj + 7yk,
C is the curve of intersection of the plane
x + z = 8
and the cylinder
x2 + y2 = 9.
You doing an MITx subject?
Just asking :)
curve is ellipse
at z = 11 and 5, y = 0 and x = +/-3
(-3 , 0 , 11)
(+3 , 0, 5)
at z = 8 , x = 0 and y = +/-3
The normal to that ellipse is in he x z plane and is perpendicular to the line joining(-3, 0 ,11)and (3,0,5)
slope of that original line dz/dx =
-6/6 = -1
remarkable
so slope of our normal is dz/dx = 1
N = 1 i + 0 j + 1k
Now maybe you can do F dot N
or rather curl F
To use Stokes' Theorem to evaluate the line integral C 𝐹 · 𝑑𝑟, we need to follow a series of steps:
Step 1: Determine the orientation and parameterization of the curve C.
C is the curve of intersection of the plane x + z = 8 and the cylinder x^2 + y^2 = 9. To find the orientation and parameterization, we can express the curve in terms of a single variable, such as t.
Given that x + z = 8 and x^2 + y^2 = 9, we can solve these equations simultaneously.
From the first equation, we have:
x = 8 - z
Substituting this into the second equation:
(8 - z)^2 + y^2 = 9
64 - 16z + z^2 + y^2 = 9
z^2 - 16z + y^2 - 55 = 0
This represents a circle in the xz-plane. Since we want C to be oriented counterclockwise as viewed from above, we can choose y = 0 and solve for z to get:
z^2 - 16z - 55 = 0
Factoring the quadratic equation:
(z - 11)(z + 5) = 0
Solving for z, we have two possible values: z = 11 and z = -5.
For z = 11, substituting back into the equation x + z = 8, we get:
x = -3
Thus, one point on the curve C is (-3, 0, 11).
For z = -5, substituting back into the equation x + z = 8, we get:
x = 13
Thus, another point on the curve C is (13, 0, -5).
Parameterizing the curve C, we can let x = 9cos(t), y = 9sin(t), and z = 8 - x.
So, the parameterization of C is:
𝑟(𝑡) = (9cos(t), 9sin(t), 8 - 9cos(t))
Step 2: Find the cross product of the parameterized curve.
Taking the derivative of 𝑟(𝑡) with respect to t, we get:
𝑟′(𝑡) = (-9sin(t), 9cos(t), 9sin(t))
Step 3: Compute the line integral.
Now, we need to compute 𝐹(𝑟(𝑡)) · (𝑟′(𝑡)) and integrate it over the parameter range of t.
𝐹(𝑟(𝑡)) = (9cos(t) * 9sin(t), 5(8 - 9cos(t)), 7(8 - 9cos(t)))
= (81sin(t)cos(t), 40 - 45cos(t), 56 - 63cos(t))
Taking the dot product of 𝐹(𝑟(𝑡)) and (𝑟′(𝑡)), we have:
𝐹(𝑟(𝑡)) · (𝑟′(𝑡)) = [81sin(t)cos(t)](-9sin(t)) + (40 - 45cos(t))9cos(t) + (56 - 63cos(t))9sin(t)
= -729sin^2(t)cos(t) - 405cos^2(t) + 360cos(t) + 504sin(t)
To evaluate the line integral, integrate 𝐹(𝑟(𝑡)) · (𝑟′(𝑡)) with respect to t over the parameter range of t.
∫(𝐹(𝑟(𝑡)) · (𝑟′(𝑡))) 𝑑𝑡 from t = 0 to t = 2π
This involves calculating the integral of each term separately and then integrating them over the given range.
The final result of the line integral will be the value obtained after evaluating the integral.