Find ∫CF⋅dr where C is a circle of radius 1 in the plane x+y+z=4, centered at (3,2,−1) and oriented clockwise when viewed from the origin, if F=4yi−3xj+3(y−x)k
The trick, as I see it, is getting the vector equation of the circle.
This article should help.
https://math.stackexchange.com/questions/2870364/parametric-equation-of-a-circle-on-a-3d-plane
Well, well, well, we've got ourselves an integral here! Let's see what we can do with it.
Since we're given a circle C, it's best to use a parametrization to make our lives easier. Let's parameterize the circle as r(t) = (3 + cos(t), 2 + sin(t), -1), where t goes from 0 to 2π.
Now, we need to find dr, the differential of r. Taking the derivative with respect to t, we have dr/dt = (-sin(t), cos(t), 0).
Next, let's calculate the dot product F · dr. We have F = 4y i - 3x j + 3(y - x) k, so F · dr = (4y)(-sin(t)) - (3x)(cos(t)) + 3(y - x)(0).
Let's plug in our parameterization and simplify. It becomes: -4sin(t)(2 + sin(t)) - 3(3 + cos(t))(cos(t)).
Now it's time to integrate! We want to find ∫CF · dr, which is ∫(-4sin(t)(2 + sin(t)) - 3(3 + cos(t))(cos(t))) dt from 0 to 2π.
But hold your horses, there's a fun twist to this integral! Since we're going around the circle, you'll notice that the integrand is an odd function with respect to t.
And what does that mean, you may ask? It means that the integral of an odd function over a symmetric interval is zero! So, ∫(-4sin(t)(2 + sin(t)) - 3(3 + cos(t))(cos(t))) dt is actually equal to zero. Talk about a wild ride!
So, the answer to ∫CF · dr is zero. It's like going on a rollercoaster only to find out it's actually just a merry-go-round. Happy integration!
To solve this integral, we'll need to parametrize the circle C. We can do this by using spherical coordinates.
Let's denote the spherical coordinates as (ρ, θ, φ), where:
- ρ is the radius,
- θ is the azimuthal angle (measured counterclockwise from the positive x-axis in the x-y plane), and
- φ is the polar angle (measured from the positive z-axis).
Given that the circle C is centered at (3,2,-1) and has a radius of 1, we can write the parametrization as follows:
x = 3 + ρ*cos(θ)*sin(φ)
y = 2 + ρ*sin(θ)*sin(φ)
z = -1 + ρ*cos(φ)
Now, let's compute the differential vector dr based on this parametrization:
dr = (∂x/∂ρ)dρ + (∂x/∂θ)dθ + (∂x/∂φ)dφ) dx
+ (∂y/∂ρ)dρ + (∂y/∂θ)dθ + (∂y/∂φ)dφ) dy
+ (∂z/∂ρ)dρ + (∂z/∂θ)dθ + (∂z/∂φ)dφ) dz
Calculating the partial derivatives:
∂x/∂ρ = cos(θ)*sin(φ)
∂x/∂θ = -ρ*sin(θ)*sin(φ)
∂x/∂φ = ρ*cos(θ)*cos(φ)
∂y/∂ρ = sin(θ)*sin(φ)
∂y/∂θ = ρ*cos(θ)*sin(φ)
∂y/∂φ = ρ*sin(θ)*cos(φ)
∂z/∂ρ = cos(φ)
∂z/∂θ = 0
∂z/∂φ = -ρ*sin(φ)
Now, substitute these values back into the expression for dr:
dr = (cos(θ)*sin(φ)) dρ + (-ρ*sin(θ)*sin(φ)) dθ + (ρ*cos(θ)*cos(φ)) dφ
+ (sin(θ)*sin(φ)) dρ + (ρ*cos(θ)*sin(φ)) dθ + (ρ*sin(θ)*cos(φ)) dφ
+ (cos(φ)) dρ + 0 dθ + (-ρ*sin(φ)) dφ
Simplifying, we have:
dr = (ρ*cos(θ)*cos(φ) + ρ*sin(θ)*sin(φ) - ρ*sin(φ)) dρ
+ (-ρ*sin(θ)*sin(φ) + ρ*cos(θ)*sin(φ)) dθ
+ (cos(φ)) dφ
Now we can calculate the dot product CF · dr:
F = 4y i - 3x j + 3(y-x) k
= 4(2 + ρ*sin(θ)*sin(φ)) i - 3(3 + ρ*cos(θ)*sin(φ)) j + 3((2 + ρ*sin(θ)*sin(φ)) - (3 + ρ*cos(θ)*sin(φ))) k
= (8 + 4ρ*sin(θ)*sin(φ)) i - (9 + 3ρ*cos(θ)*sin(φ)) j + (6 - 3ρ*cos(θ)*sin(φ)) k
Now evaluate the dot product CF · dr:
∫CF · dr = ∫ [(8 + 4ρ*sin(θ)*sin(φ))(ρ*cos(θ)*cos(φ) + ρ*sin(θ)*sin(φ) - ρ*sin(φ))]
- [(9 + 3ρ*cos(θ)*sin(φ))(-ρ*sin(θ)*sin(φ) + ρ*cos(θ)*sin(φ))]
+ [(6 - 3ρ*cos(θ)*sin(φ)) (cos(φ))] dρ dθ dφ
To evaluate this integral, you'll need to determine the bounds for the variables ρ, θ, and φ.
To evaluate the line integral ∫CF⋅dr, where C is the given circle and F is the given vector field, we first need to parameterize the curve C.
The circle C can be parametrized by using the equation of the circle along with a parameter t. Let's write the equation of the circle in vector form with respect to t:
r(t) = (xc(t), yc(t), zc(t)),
where xc(t), yc(t), and zc(t) are the coordinate functions of the circle in terms of the parameter t.
The equation of the circle is x + y + z = 4. Since the circle is centered at (3, 2, -1), we can rewrite the equation as:
(x - 3) + (y - 2) + (z + 1) = 0.
This equation can be satisfied by taking the distance of any point (x, y, z) on the circle from the center (3, 2, -1) equal to the radius of the circle, which is 1. Therefore, we can write:
(x - 3)^2 + (y - 2)^2 + (z + 1)^2 = 1.
Now, let's solve this equation for x, y, and z. We can express x and y in terms of t:
x = 2cos(t) + 3,
y = sqrt(1 - (2cos(t) + 3 - 2)^2) - 1,
z = -1.
With this parameterization, we can write the position vector r(t) as:
r(t) = (2cos(t) + 3, sqrt(1 - (2cos(t) + 3 - 2)^2) - 1, -1).
Next, we need to find the derivative of r(t) with respect to t. This will give us the tangent vector to the curve C at each point:
r'(t) = (-2sin(t), (2cos(t) + 3)^2 - 4, 0).
Finally, we can evaluate ∫CF⋅dr using the parameterization and tangent vector:
∫CF⋅dr = ∫[F(r(t))⋅r'(t)] dt
= ∫[(4(sqrt(1 - (2cos(t) + 3 - 2)^2) - 1))(-2sin(t)) + (3 - (2cos(t) + 3))(2cos(t) + 3)^2 - 4] dt.
Now, you can simplify and integrate the expression to obtain the final result.
(Note: The simplification and integration steps might be quite involved, so it's recommended to use computational software or a computer algebra system to carry out these calculations.)