Mixing
Problem arising while calculating surface integral by taking projection.
$begingroup$
I was asked to verify the divergence theorem for $$vec{A}=4xhat{i}-2y^2hat{j}+z^2hat{k}$$
taken over the region bounded by $$x^2+y^2=4,z=0$$ and $$z=3$$.
One part ($$iiintnabla.vec{A}dV$$) is trivial . Now for the other part, when I tried to solve by taking
projection on the $$xz$$ plane I got one value and on making the substitution $$x=r costheta, y=r sintheta$$ I got a different value. Can someone please explain the reason for the difference?
surface-integrals divergence
asked Jan 20 at 10:51
user566574user566574
114
$endgroup$
add a comment |
$begingroup$
I was asked to verify the divergence theorem for $$vec{A}=4xhat{i}-2y^2hat{j}+z^2hat{k}$$
taken over the region bounded by $$x^2+y^2=4,z=0$$ and $$z=3$$.
One part ($$iiintnabla.vec{A}dV$$) is trivial . Now for the other part, when I tried to solve by taking
projection on the $$xz$$ plane I got one value and on making the substitution $$x=r costheta, y=r sintheta$$ I got a different value. Can someone please explain the reason for the difference?
surface-integrals divergence
asked Jan 20 at 10:51
user566574user566574
114
$endgroup$
add a comment |
$begingroup$
I was asked to verify the divergence theorem for $$vec{A}=4xhat{i}-2y^2hat{j}+z^2hat{k}$$
taken over the region bounded by $$x^2+y^2=4,z=0$$ and $$z=3$$.
One part ($$iiintnabla.vec{A}dV$$) is trivial . Now for the other part, when I tried to solve by taking
projection on the $$xz$$ plane I got one value and on making the substitution $$x=r costheta, y=r sintheta$$ I got a different value. Can someone please explain the reason for the difference?
surface-integrals divergence
asked Jan 20 at 10:51
user566574user566574
114
$endgroup$
I was asked to verify the divergence theorem for $$vec{A}=4xhat{i}-2y^2hat{j}+z^2hat{k}$$
taken over the region bounded by $$x^2+y^2=4,z=0$$ and $$z=3$$.
One part ($$iiintnabla.vec{A}dV$$) is trivial . Now for the other part, when I tried to solve by taking
projection on the $$xz$$ plane I got one value and on making the substitution $$x=r costheta, y=r sintheta$$ I got a different value. Can someone please explain the reason for the difference?
surface-integrals divergence
surface-integrals divergence
asked Jan 20 at 10:51
user566574user566574
114
asked Jan 20 at 10:51
user566574user566574
114
asked Jan 20 at 10:51
user566574user566574
114
asked Jan 20 at 10:51
user566574user566574
114
asked Jan 20 at 10:51
user566574user566574
114
114
add a comment |
add a comment |
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