Set up a double integral for calculating the flux of the vector field F⃗ (r⃗ )=r⃗ , where r⃗ =⟨x,y,z⟩, through the part of the upward oriented surface z=9(x^2+y^2) that lies above the disk x^2+y^2≤25.
Why did the vector field get invited to the party? Because it knows how to "flux" it up!
Alright, let's get into the mathematical shenanigans. We want to calculate the flux of the vector field F⃗ (r⃗ )=r⃗ through the surface z=9(x^2+y^2), which is bounded by the disk x^2+y^2≤25. To set up the double integral, we'll need to parametrize the surface.
Since we have z=9(x^2+y^2), we can let x=r*cos(θ) and y=r*sin(θ), where r is the radius and θ is the angle. This parametrization maps the disk x^2+y^2≤25 onto the surface z=9r^2.
The surface element dS can be calculated as dS = |∂r/∂x x ∂r/∂y| dA, where ∂r/∂x and ∂r/∂y are the partial derivatives of r = ⟨x,y,9(x^2+y^2)⟩ with respect to x and y. Since r only depends on x and y, these partial derivatives are constants and we can calculate their magnitudes.
∂r/∂x = ⟨1,0,18x⟩ and ∂r/∂y = ⟨0,1,18y⟩. The magnitude of ∂r/∂x is √(1^2 + 0^2 + (18x)^2) = √(1 + 324x^2) and the magnitude of ∂r/∂y is √(0^2 + 1^2 + (18y)^2) = √(1 + 324y^2).
Now we have all the ingredients to set up the integral: ∬ F⃗ · dS = ∬ F⃗ · (∂r/∂x x ∂r/∂y) dA.
Using the parametrization x=r*cos(θ) and y=r*sin(θ), the integral becomes ∬ (⟨r*cos(θ), r*sin(θ), 9r^2⟩) · (∂r/∂x x ∂r/∂y) dA.
The disk x^2+y^2≤25 maps to r ranging from 0 to 5 and θ ranging from 0 to 2π.
So, the final setup for the double integral is:
∬ (r*cos(θ), r*sin(θ), 9r^2) · (∂r/∂x x ∂r/∂y) dA
over the region r: 0 to 5 and θ: 0 to 2π.
I hope the flux calculation doesn't give you a headache! Don't forget to bring a calculator to the party!
To set up a double integral for calculating the flux of the vector field F⃗ (r⃗ )=r⃗ through the surface described, we need to parameterize the surface and calculate the dot product of the field and the normal vector to the surface.
First, let's find the normal vector to the surface. The surface equation is z = 9(x^2 + y^2), which represents a paraboloid that opens upward. The normal vector can be found by taking the gradient of the surface equation.
∇z = ⟨∂/∂x, ∂/∂y, -1⟩
Next, we need to parameterize the surface. Since the surface lies above the disk x^2 + y^2 ≤ 25, we can use polar coordinates.
Let x = r*cosθ and y = r*sinθ, where r ranges from 0 to 5 (since the radius of the disk is 5) and θ ranges from 0 to 2π (for a complete revolution).
Substituting these values in the surface equation, we get:
z = 9(r^2*cos^2θ + r^2*sin^2θ) = 9r^2.
Now, let's calculate the dot product of the vector field F⃗ and the normal vector ∇z.
F⃗ dot ∇z = ⟨r*cosθ, r*sinθ, 9r²⟩ dot ⟨∂/∂x, ∂/∂y, -1⟩
= r*cosθ * ∂/∂x + r*sinθ * ∂/∂y - 9r² * ∂/∂z.
Next, we need to calculate the magnitude of the normal vector to find the area element of the surface.
|∇z| = √(∂z/∂x)^2 + (∂z/∂y)^2 + (-1)^2
= √(2x^2 + 2y^2 + 1)
= √(2r^2 + 1).
Finally, we set up the double integral to calculate the flux:
Flux = ∬S (F⃗ dot ∇z) |∇z| dA
= ∬R (r*cosθ * ∂/∂x + r*sinθ * ∂/∂y - 9r² * ∂/∂z) √(2r^2 + 1) dr dθ.
Here, R represents the region in the xy-plane defined by the disk x^2 + y^2 ≤ 25, and dA represents the area element in the xy-plane.
Therefore, the double integral to calculate the flux is:
Flux = ∫(θ=0 to 2π) ∫(r=0 to 5) (r*cosθ * ∂/∂x + r*sinθ * ∂/∂y - 9r² * ∂/∂z) √(2r^2 + 1) dr dθ.
To set up a double integral for calculating the flux of the vector field F⃗ (r⃗ ) = r⃗ through the specified surface, we need to follow these steps:
Step 1: Determine the bounds of integration.
Step 2: Set up the integral.
Let's go through each step:
Step 1: Determine the bounds of integration.
We have a surface defined by the equation z = 9(x^2 + y^2), which represents a paraboloid centered at the origin with a radius of 5 units. The surface lies above the disk x^2 + y^2 ≤ 25, which is a circle centered at the origin with a radius of 5 units.
To set up the double integral, we need to determine the bounds of integration for x and y. Since the surface lies above the disk, for any given x, the values of y will only be within the range of the circle centered at the origin.
For x, we can choose the range [-5, 5] because the radius of the disk is 5 units.
For y, we can choose the range [-√(25 - x^2), √(25 - x^2)] based on the equation of the circle.
Step 2: Set up the integral.
The flux of a vector field through a surface is given by the double integral of the dot product of the vector field F⃗ (r⃗ ) and the normal vector dS⃗ over the surface.
The normal vector to the surface z = 9(x^2 + y^2) is given by (-∂z/∂x, -∂z/∂y, 1) = (-18x, -18y, 1).
The dot product of the vector field F⃗ (r⃗ ) = r⃗ and the normal vector dS⃗ is then (-18x, -18y, 1)·(x, y, 9(x^2 + y^2)).
The flux integral can be set up as:
∬(S) (-18x, -18y, 1)·(x, y, 9(x^2 + y^2)) dS
Using the definition of dot product, this simplifies to:
∬(S) (-18x^2 - 18y^2 + 9(x^2 + y^2)) dS
Now, since we are integrating over a surface parameterized by x and y, we can express the surface area element dS in terms of the partial derivatives of the parameterization.
dS = √(1 + (∂z/∂x)^2 + (∂z/∂y)^2) dxdy
Substituting the values of the partial derivatives of z with respect to x and y, we get:
dS = √(1 + 324x^2 + 324y^2) dxdy
Finally, combining this with the previous expression for the flux integral, we have:
∬(S) (-18x^2 - 18y^2 + 9(x^2 + y^2)) √(1 + 324x^2 + 324y^2) dxdy
And this is the setup for the double integral to calculate the flux of the vector field through the specified surface.