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The divergence theorem of the equation S(2x+2y+z2)dS exists 4/3 π.
What is meant by divergence theorem?
Let S be the boundary surface of E with the positive (outward) orientation, and let E be a straightforward solid region. Let F be a vector field with continuous partial derivatives on an open region containing E for each of its component functions. Then
[tex]$\iint_S F \cdot d S=\iint_S F \cdot n d S=\iiint_E div Fdv$[/tex],
where n be the outward normal of S.
Let the given equation be
The sphere S is x² + y² + z² = 1
By using concept,
[tex]$\iint_S F \cdot n d S=\iint_S\left(2 x+2 y+z^2\right) d S$[/tex]
Therefore,
[tex]$F \cdot n=2 x+2 y+z^2$[/tex]
Given that the sphere S is x² + y² + z² = 1
For S, n will be
[tex]$n=\frac{x i+y j+z k}{\sqrt{x^2+y^2+z^2}}$[/tex]
After substituting the value of x² + y² + z² = 1
n = x i + y j + z k
Therefore,
F(xi + yj + zk) = 2x + 2y + z²
Consider F = Pi + Qj + Rk
[tex]${data-answer}amp; (P i+Q j+R k) \cdot(x i+y j+z k)=2 x+2 y+z^2 \\[/tex]
Px + Qy + Rz = 2x + 2y + z²
By comparing the values, P = 2, Q = 2, R = z
So, F becomes
F = 2i + 2j + zk
By using the definition of Divergence,
[tex]${div} F=\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) \cdot(2 i+2 j+z k) \\[/tex]
[tex]${data-answer}amp; =\frac{\partial}{\partial x}(2 i+2 j+z k)+\frac{\partial}{\partial y}(2 i+2 j+z k)+\frac{\partial}{\partial z}(2 i+2 j+z k) \\[/tex]
= 0 + 0 + 1
div F = 1
By using Divergence Theorem,
[tex]$ \iint_S F \cdot n d S=\iiint_E {div} F d V \\[/tex]
[tex]${data-answer}amp; \iint_S\left(2 x+2 y+z^2\right) d S=\iiint_E 1 d V[/tex]
Since [tex]$\iiint_E 1 d V$[/tex] is the volume of sphere with radius 1
[tex]${data-answer}amp; \iint_S\left(2 x+2 y+z^2\right) d S=\text { Volume of } E \\[/tex]
[tex]${data-answer}amp; =\frac{4}{3} \pi(1)^3=\frac{4}{3} \pi[/tex]
Thus,
[tex]$\iint_S\left(2 x+2 y+z^2\right) d S=\frac{4}{3} \pi$[/tex]
Therefore, the divergence theorem of the equation S(2x+2y+z2)dS exists 4/3 π.
The complete question is:
Use the Divergence Theorem to evaluate S(2x+2y+z2)dS where S is the sphere x2+y2+z2=1.
To learn more about Divergence Theorem refer to:
https://brainly.com/question/17177764
#SPJ4
The divergence theorem of the equation S(2x+2y+z2)dS exists 4/3 π.
What is meant by divergence theorem?
Let S be the boundary surface of E with the positive (outward) orientation, and let E be a straightforward solid region. Let F be a vector field with continuous partial derivatives on an open region containing E for each of its component functions. Then
[tex]\iint_S F \cdot d S=\iint_S F \cdot n d S=\iiint_E d i v F d v[/tex]
where n be the outward normal of S.
Let the given equation be
The sphere S is x² + y² + z² = 1
By using concept,
[tex]\iint_S F \cdot n d S=\iint_S\left(2 x+2 y+z^2\right) d S[/tex]
Therefore,
[tex]\mathrm{F} \cdot n=2 x+2 y+z^2[/tex]
Given that the sphere S is x² + y² + z² = 1
For S, n will be
[tex]n=\frac{x i+y j+z k}{\sqrt{x^2+y^2+z^2}}[/tex]
After substituting the value of x² + y² + z² = 1
n = x i + y j + z k
Therefore,
F(xi + yj + zk) = 2x + 2y + z²
Consider F = Pi + Qj + Rk
[tex](P i+Q j+R k) \cdot(x i+y j+z k)=2 x+2 y+z^2[/tex]
Px + Qy + Rz = 2x + 2y + z²
By using the definition of Divergence,
= 0 + 0 + 1
div F = 1
By using Divergence Theorem,
[tex]\begin{aligned}& \iint_S F \cdot n d S=\iiint_E d i v F d V \\& \iint_S\left(2 x+2 y+z^2\right) d S=\iiint_E 1 d V\end{aligned}[/tex]
Since[tex]\iiint_E 1 d V[/tex]is the volume of sphere with radius 1[tex]$\begin{aligned}& \iint_S\left(2 x+2 y+z^2\right) d S=\text { Volume of } E \\& =\frac{4}{3} \pi(1)^3=\frac{4}{3} \pi\end{aligned}$Thus,\iint_S\left(2 x+2 y+z^2\right) d S=\frac{4}{3} \pi[/tex]
Therefore, the divergence theorem of the equation S(2x+2y+z2)dS exists 4/3 π.
Use the Divergence Theorem to evaluate S(2x+2y+z2)dS where S is the sphere x2+y2+z2=1.
To learn more about Divergence Theorem visit:
brainly.com/question/17177764
#SPJ4
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