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Calculate λSFndS\iint_{\lambda S} \mathbf{F} \cdot \mathbf{n} d S for each of the following. Looked at the right way, all are quite easy and some are even trivial.

(a) F=(2x+yz)i+3yj+z2k\mathbf{F}=(2 x+y z) \mathbf{i}+3 y \mathbf{j}+z^2 \mathbf{k}; SS is the solid sphere x2+y2+z21x^2+y^2+z^2 \leq 1.

(b) F=(x2+y2+z2)5/3(xi+yj+zk);S\mathbf{F}=\left(x^2+y^2+z^2\right)^{5 / 3}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) ; S as in part (a).

(c) F=x2i+y2j+z2k\mathbf{F}=x^2 \mathbf{i}+y^2 \mathbf{j}+z^2 \mathbf{k}; S is the solid sphere (x2)2+y2+z21(x-2)^2+y^2+z^2 \leq 1.

(d) F=x2i;S\mathbf{F}=x^2 \mathbf{i} ; S is the cube 0x1,0y1,0z10 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1.

(e) F=(x+z)i+(y+x)j+(z+y)k;S\mathbf{F}=(x+z) \mathbf{i}+(y+x) \mathbf{j}+(z+y) \mathbf{k} ; S is the tetrahedron cut from the first octant by the plane 3x+4y+2z=123 x+4 y+2 z=12.

(f) F=x3i+y3j+z3k;S\mathbf{F}=x^3 \mathbf{i}+y^3 \mathbf{j}+z^3 \mathbf{k} ; S as in part (a).

(g) F=(xi+yj)ln(x2+y2)\mathbf{F}=(x \mathbf{i}+y \mathbf{j}) \ln \left(x^2+y^2\right); S is the solid cylinder x2+y24,0z2x^2+y^2 \leq 4,0 \leq z \leq 2.


Answered 2 years ago
Answered 2 years ago
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Let's find the divergence of the given vector field. Since F(x,y,z)=(2x+yz)i+3yj+z2k{\bf F}(x,y,z)=(2x+yz){\bf i}+3y{\bf j}+z^2{\bf k} then P(x,y,z)=2x+yz,Q(x,y,z)=3yP(x,y,z)=2x+yz, Q(x,y,z)=3y and R(x,y,z)=z2R(x,y,z)=z^2 and therefore:

F=Px+Qy+Rz=(2x+yz)x+(3y)y+(z2)z=2+3+2z=5+2z\begin{aligned} \nabla\cdot{\bf F}&=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}\\ &=\dfrac{\partial (2x+yz)}{\partial x}+\dfrac{\partial (3y)}{\partial y}+\dfrac{\partial (z^2)}{\partial z}\\ &=2+3+2z\\ &=5+2z \end{aligned}

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