Evaluate the vector field $(xz^2, z^3, z(x+y))$ over an ellipsoid
$begingroup$
I would like to evaluate the following vector function
$$(xz^2, z^3, z(x+y))$$
over the surface of the ellipsoid: $$x^2/4 + y^2/2 + z^2 = 1$$
I thought of using the divergence theorem to instead evaluate the triple integral:
$$iiint (z^2+x+y)dV$$ but even though this looks much simpler, I still get bogged down by the computation relatively quickly. I tried changing the variables so that we get a sphere instead: $u = x/2$, $v = y/sqrt(2)$ and z = z, then followed by a spherical change of variables. Is this the best way to approach the problem, or is there another way that will make the computation much simpler?
multivariable-calculus vector-fields
asked Dec 5 '18 at 2:00
mmmmommmmo
1087
$endgroup$
add a comment |
$begingroup$
I would like to evaluate the following vector function
$$(xz^2, z^3, z(x+y))$$
over the surface of the ellipsoid: $$x^2/4 + y^2/2 + z^2 = 1$$
I thought of using the divergence theorem to instead evaluate the triple integral:
$$iiint (z^2+x+y)dV$$ but even though this looks much simpler, I still get bogged down by the computation relatively quickly. I tried changing the variables so that we get a sphere instead: $u = x/2$, $v = y/sqrt(2)$ and z = z, then followed by a spherical change of variables. Is this the best way to approach the problem, or is there another way that will make the computation much simpler?
multivariable-calculus vector-fields
asked Dec 5 '18 at 2:00
mmmmommmmo
1087
$endgroup$
add a comment |
$begingroup$
I would like to evaluate the following vector function
$$(xz^2, z^3, z(x+y))$$
over the surface of the ellipsoid: $$x^2/4 + y^2/2 + z^2 = 1$$
I thought of using the divergence theorem to instead evaluate the triple integral:
$$iiint (z^2+x+y)dV$$ but even though this looks much simpler, I still get bogged down by the computation relatively quickly. I tried changing the variables so that we get a sphere instead: $u = x/2$, $v = y/sqrt(2)$ and z = z, then followed by a spherical change of variables. Is this the best way to approach the problem, or is there another way that will make the computation much simpler?
multivariable-calculus vector-fields
asked Dec 5 '18 at 2:00
mmmmommmmo
1087
$endgroup$
I would like to evaluate the following vector function
$$(xz^2, z^3, z(x+y))$$
over the surface of the ellipsoid: $$x^2/4 + y^2/2 + z^2 = 1$$
I thought of using the divergence theorem to instead evaluate the triple integral:
$$iiint (z^2+x+y)dV$$ but even though this looks much simpler, I still get bogged down by the computation relatively quickly. I tried changing the variables so that we get a sphere instead: $u = x/2$, $v = y/sqrt(2)$ and z = z, then followed by a spherical change of variables. Is this the best way to approach the problem, or is there another way that will make the computation much simpler?
multivariable-calculus vector-fields
multivariable-calculus vector-fields
asked Dec 5 '18 at 2:00
mmmmommmmo
1087
asked Dec 5 '18 at 2:00
mmmmommmmo
1087
edited Dec 5 '18 at 15:29
mmmmo
asked Dec 5 '18 at 2:00
mmmmommmmo
1087
asked Dec 5 '18 at 2:00
mmmmommmmo
1087
asked Dec 5 '18 at 2:00
mmmmommmmo
1087
1087
add a comment |
add a comment |
1 Answer
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I would first split the integral in three parts, one for $z^2$, one for $x$, and one for $y$. It is easy to show that the last two integrals are zero. Now you do your change of variables to spherical coordinates, just as you proposed. It should be relatively easy to solve this integral, since you can write it as a product of three independent integrals. Use $z=rcostheta$
answered Dec 5 '18 at 2:31
AndreiAndrei
11.7k21026
$endgroup$
$begingroup$
This is correct, I did not think to check if any part of the integral went to 0 before doing my change of variables. For anyone who finds this problem in the future, I will leave the final answer here: $(8pi)(sqrt(2)/15)$
$endgroup$
– mmmmo
Dec 5 '18 at 15:28
add a comment |
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1 Answer
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active
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1 Answer
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active
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active
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$begingroup$
I would first split the integral in three parts, one for $z^2$, one for $x$, and one for $y$. It is easy to show that the last two integrals are zero. Now you do your change of variables to spherical coordinates, just as you proposed. It should be relatively easy to solve this integral, since you can write it as a product of three independent integrals. Use $z=rcostheta$
answered Dec 5 '18 at 2:31
AndreiAndrei
11.7k21026
$endgroup$
$begingroup$
This is correct, I did not think to check if any part of the integral went to 0 before doing my change of variables. For anyone who finds this problem in the future, I will leave the final answer here: $(8pi)(sqrt(2)/15)$
$endgroup$
– mmmmo
Dec 5 '18 at 15:28
add a comment |
$begingroup$
I would first split the integral in three parts, one for $z^2$, one for $x$, and one for $y$. It is easy to show that the last two integrals are zero. Now you do your change of variables to spherical coordinates, just as you proposed. It should be relatively easy to solve this integral, since you can write it as a product of three independent integrals. Use $z=rcostheta$
answered Dec 5 '18 at 2:31
AndreiAndrei
11.7k21026
$endgroup$
$begingroup$
This is correct, I did not think to check if any part of the integral went to 0 before doing my change of variables. For anyone who finds this problem in the future, I will leave the final answer here: $(8pi)(sqrt(2)/15)$
$endgroup$
– mmmmo
Dec 5 '18 at 15:28
add a comment |
$begingroup$
I would first split the integral in three parts, one for $z^2$, one for $x$, and one for $y$. It is easy to show that the last two integrals are zero. Now you do your change of variables to spherical coordinates, just as you proposed. It should be relatively easy to solve this integral, since you can write it as a product of three independent integrals. Use $z=rcostheta$
answered Dec 5 '18 at 2:31
AndreiAndrei
11.7k21026
$endgroup$
I would first split the integral in three parts, one for $z^2$, one for $x$, and one for $y$. It is easy to show that the last two integrals are zero. Now you do your change of variables to spherical coordinates, just as you proposed. It should be relatively easy to solve this integral, since you can write it as a product of three independent integrals. Use $z=rcostheta$
answered Dec 5 '18 at 2:31
AndreiAndrei
11.7k21026
answered Dec 5 '18 at 2:31
AndreiAndrei
11.7k21026
answered Dec 5 '18 at 2:31
AndreiAndrei
11.7k21026
answered Dec 5 '18 at 2:31
AndreiAndrei
11.7k21026
11.7k21026
$begingroup$
This is correct, I did not think to check if any part of the integral went to 0 before doing my change of variables. For anyone who finds this problem in the future, I will leave the final answer here: $(8pi)(sqrt(2)/15)$
$endgroup$
– mmmmo
Dec 5 '18 at 15:28
add a comment |
$begingroup$
This is correct, I did not think to check if any part of the integral went to 0 before doing my change of variables. For anyone who finds this problem in the future, I will leave the final answer here: $(8pi)(sqrt(2)/15)$
$endgroup$
– mmmmo
Dec 5 '18 at 15:28
$begingroup$
This is correct, I did not think to check if any part of the integral went to 0 before doing my change of variables. For anyone who finds this problem in the future, I will leave the final answer here: $(8pi)(sqrt(2)/15)$
$endgroup$
– mmmmo
Dec 5 '18 at 15:28
$begingroup$
This is correct, I did not think to check if any part of the integral went to 0 before doing my change of variables. For anyone who finds this problem in the future, I will leave the final answer here: $(8pi)(sqrt(2)/15)$
$endgroup$
– mmmmo
Dec 5 '18 at 15:28
add a comment |
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