Calculate the exact flux across the surface

Calculate the exact flux across the surface, Jfe(x,y,z)edS for the vector field F(x,v,2)-%+li+ x-ljtzk 3x 4y where S is the closed surface consisting of two parts: Paraboloid: Disk: x^2 + y^2 + 2 = 9, x + y < 9 2z = 0

The Correct Answer and Explanation is:

To calculate the flux across the surface SSS for the vector field F(x,y,z)\mathbf{F}(x, y, z)F(x,y,z), we need to apply the surface integral. The vector field is given as:F(x,y,z)=⟨x−y+z,x+y,3x+4y⟩\mathbf{F}(x, y, z) = \langle x – y + z, x + y, 3x + 4y \rangleF(x,y,z)=⟨x−y+z,x+y,3x+4y⟩

We are asked to compute the flux ∬SF⋅n dS\iint_S \mathbf{F} \cdot \mathbf{n} \, dS∬S​F⋅ndS, where n\mathbf{n}n is the unit normal vector to the surface SSS and dSdSdS represents the surface element.

Step 1: Surface Breakdown

The surface SSS consists of two parts:

1. Paraboloid: x2+y2+2=9x^2 + y^2 + 2 = 9×2+y2+2=9, this represents a paraboloid opening upward.
2. Disk: 2z=02z = 02z=0, which is a disk at z=0z = 0z=0.

Thus, the surface consists of the upper half of a paraboloid, and the flat disk at z=0z = 0z=0.

Step 2: Divergence Theorem

Since the surface SSS is closed, we can use the Divergence Theorem to simplify our calculations. The Divergence Theorem states:∬SF⋅n dS=∭V(∇⋅F) dV\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_V (\nabla \cdot \mathbf{F}) \, dV∬S​F⋅ndS=∭V​(∇⋅F)dV

where VVV is the volume enclosed by SSS, and ∇⋅F\nabla \cdot \mathbf{F}∇⋅F is the divergence of the vector field F\mathbf{F}F.

Step 3: Calculate the Divergence of F\mathbf{F}F

The divergence ∇⋅F\nabla \cdot \mathbf{F}∇⋅F is computed by taking the partial derivatives of each component of F\mathbf{F}F:F=⟨x−y+z,x+y,3x+4y⟩\mathbf{F} = \langle x – y + z, x + y, 3x + 4y \rangleF=⟨x−y+z,x+y,3x+4y⟩∇⋅F=∂∂x(x−y+z)+∂∂y(x+y)+∂∂z(3x+4y)\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x – y + z) + \frac{\partial}{\partial y}(x + y) + \frac{\partial}{\partial z}(3x + 4y)∇⋅F=∂x∂​(x−y+z)+∂y∂​(x+y)+∂z∂​(3x+4y)

Calculating the partial derivatives:∂∂x(x−y+z)=1\frac{\partial}{\partial x}(x – y + z) = 1∂x∂​(x−y+z)=1∂∂y(x+y)=1\frac{\partial}{\partial y}(x + y) = 1∂y∂​(x+y)=1∂∂z(3x+4y)=0\frac{\partial}{\partial z}(3x + 4y) = 0∂z∂​(3x+4y)=0

Thus, the divergence is:∇⋅F=1+1+0=2\nabla \cdot \mathbf{F} = 1 + 1 + 0 = 2∇⋅F=1+1+0=2

Step 4: Set Up the Volume Integral

Now we can set up the volume integral. Since ∇⋅F=2\nabla \cdot \mathbf{F} = 2∇⋅F=2, we need to integrate over the volume VVV enclosed by the surface. The volume is the region bounded by the paraboloid and the disk at z=0z = 0z=0.

The equation of the paraboloid is x2+y2=9−2zx^2 + y^2 = 9 – 2zx2+y2=9−2z. We will convert the region to cylindrical coordinates to simplify the integration.

In cylindrical coordinates:

• x=rcos⁡θx = r \cos \thetax=rcosθ
• y=rsin⁡θy = r \sin \thetay=rsinθ
• z=zz = zz=z
• The volume element dV=r dz dr dθdV = r \, dz \, dr \, d\thetadV=rdzdrdθ

The bounds for the integration are:

• rrr ranges from 0 to 3 (since x2+y2=9x^2 + y^2 = 9×2+y2=9)
• θ\thetaθ ranges from 0 to 2π2\pi2π
• zzz ranges from 0 to 9−r22\frac{9 – r^2}{2}29−r2​ (from the equation of the paraboloid)

Thus, the flux is:∬SF⋅n dS=∭V2 r dz dr dθ\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_V 2 \, r \, dz \, dr \, d\theta∬S​F⋅ndS=∭V​2rdzdrdθ

Step 5: Perform the Integration

Now, integrate over the volume VVV:∫02π∫03∫09−r222r dz dr dθ\int_0^{2\pi} \int_0^3 \int_0^{\frac{9 – r^2}{2}} 2r \, dz \, dr \, d\theta∫02π​∫03​∫029−r2​​2rdzdrdθ

First, integrate with respect to zzz:∫09−r222r dz=2r(9−r22)=r(9−r2)\int_0^{\frac{9 – r^2}{2}} 2r \, dz = 2r \left( \frac{9 – r^2}{2} \right) = r(9 – r^2)∫029−r2​​2rdz=2r(29−r2​)=r(9−r2)

Next, integrate with respect to rrr:∫03r(9−r2) dr=∫03(9r−r3) dr\int_0^3 r(9 – r^2) \, dr = \int_0^3 (9r – r^3) \, dr∫03​r(9−r2)dr=∫03​(9r−r3)dr

This is a standard polynomial integral:∫039r dr=[9r22]03=9×92=40.5\int_0^3 9r \, dr = \left[ \frac{9r^2}{2} \right]_0^3 = \frac{9 \times 9}{2} = 40.5∫03​9rdr=[29r2​]03​=29×9​=40.5∫03r3 dr=[r44]03=814=20.25\int_0^3 r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^3 = \frac{81}{4} = 20.25∫03​r3dr=[4r4​]03​=481​=20.25

Thus:∫03(9r−r3) dr=40.5−20.25=20.25\int_0^3 (9r – r^3) \, dr = 40.5 – 20.25 = 20.25∫03​(9r−r3)dr=40.5−20.25=20.25

Finally, integrate with respect to θ\thetaθ:∫02π20.25 dθ=20.25×2π=40.5π\int_0^{2\pi} 20.25 \, d\theta = 20.25 \times 2\pi = 40.5\pi∫02π​20.25dθ=20.25×2π=40.5π

Step 6: Final Answer

Thus, the flux across the surface is:40.5π\boxed{40.5\pi}40.5π​