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91Ó°ÊÓ

Evaluate the integral:

∬SF(x,y,z)·nds

F(x,y,z)=sinycoszi+yz2j+zx2kand S is the surface of the region W bounded by the paraboloidy=x2+z2and the planesy=1andy=4

Short Answer

Expert verified

The integral is∬SF(x,y,z)·ndS=21π2

Step by step solution

01

Given Information

It is given that F(x,y,z)=sinycoszi+yz2j+zx2k

We need to evaluate∬SF(x,y,z)·ndS

02

Use of Divergence Theorem to evaluate the Integral

Theorem states that:

Let W be a bounded region in R3whose boundary S is a smooth or piecewise-smooth closed oriented surface. If a vector field role="math" F(x,y,z)is defined on an open region containing W, then,

∬SF(x,y,z)·ndS=∭WdivF(x,y,z)dV

n is outwards unit normal vector.

03

Finding the Divergence

The vector field is F(x,y,z)=sinycoszi+yz2j+zx2k

It is written as:

F(x,y,z)=F1(x,y,z)i+F2(x,y,z)j+F3(x,y,z)k

Finding the divergence

divF(x,y,z)=∂∂xi+∂∂yj+∂∂zk·F1i+F2j+F3k

=∂F1∂x+∂F2∂y+∂F3∂z

Hence, the divergence is

divF(x,y,z)=∂∂xi+∂∂yj+∂∂zk·sinycoszi+yz2j+zx2k

=∂∂x(sinycosz)+∂∂yyz2+∂∂zzx2

=0+z2+x2

=x2+z2

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