Get the information you need from a community of experts on IDNLearn.com. Get accurate and comprehensive answers from our network of experienced professionals.
Sagot :
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
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
