calculate the surface integral in the upper hemisphere
Calculate the surface integral $f(x,y,z)=x^2+y^2+z^2$ in the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$
I tried to compute the value of the surface integral $iint_S{F.n} dS$ with the explicit representation z=$sqrt{-x^2-y^2-2z}$ but the term $2z$ brings me conflict.
I don´t know if it should be done like this or it is convenient to change coordinates.
I would appreciate someone helping me. Since it is for a job that is delivered today and I am stuck.
calculus surface-integrals stokes-theorem
asked Nov 27 at 11:44
F.A. Alday
1
add a comment |
Calculate the surface integral $f(x,y,z)=x^2+y^2+z^2$ in the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$
I tried to compute the value of the surface integral $iint_S{F.n} dS$ with the explicit representation z=$sqrt{-x^2-y^2-2z}$ but the term $2z$ brings me conflict.
I don´t know if it should be done like this or it is convenient to change coordinates.
I would appreciate someone helping me. Since it is for a job that is delivered today and I am stuck.
calculus surface-integrals stokes-theorem
asked Nov 27 at 11:44
F.A. Alday
1
add a comment |
Calculate the surface integral $f(x,y,z)=x^2+y^2+z^2$ in the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$
I tried to compute the value of the surface integral $iint_S{F.n} dS$ with the explicit representation z=$sqrt{-x^2-y^2-2z}$ but the term $2z$ brings me conflict.
I don´t know if it should be done like this or it is convenient to change coordinates.
I would appreciate someone helping me. Since it is for a job that is delivered today and I am stuck.
calculus surface-integrals stokes-theorem
asked Nov 27 at 11:44
F.A. Alday
1
Calculate the surface integral $f(x,y,z)=x^2+y^2+z^2$ in the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$
I tried to compute the value of the surface integral $iint_S{F.n} dS$ with the explicit representation z=$sqrt{-x^2-y^2-2z}$ but the term $2z$ brings me conflict.
I don´t know if it should be done like this or it is convenient to change coordinates.
I would appreciate someone helping me. Since it is for a job that is delivered today and I am stuck.
calculus surface-integrals stokes-theorem
calculus surface-integrals stokes-theorem
asked Nov 27 at 11:44
F.A. Alday
1
asked Nov 27 at 11:44
F.A. Alday
1
asked Nov 27 at 11:44
F.A. Alday
1
asked Nov 27 at 11:44
F.A. Alday
1
asked Nov 27 at 11:44
F.A. Alday
1
1
add a comment |
add a comment |
1 Answer
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Hoping that by "upper" hemisphere, you mean "with respect to $z$:
Let
$$
Q = int_S x^2+y^2+z^2 ~ dA
$$, where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.
Substituting $u = z-1; u+1 = z; du = dz$, we get
$$
Q = int_T x^2+y^2+(u+1)^2 ~ dA
$$, where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
$$
Q = int_U x^2+y^2+(z+1)^2 ~ dA,
$$
where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.
Expanding the integrand, we get
begin{align}
Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
&= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 2 ~dA+ int_U 2z~dA\
&= 4pi+ 2int_U z~dA
end{align}
The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.
Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
$$
(x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
$$
is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.
So the overall value of your integral is $6pi$.
answered Nov 27 at 13:42
John Hughes
62.3k24090
add a comment |
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1 Answer
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1 Answer
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Hoping that by "upper" hemisphere, you mean "with respect to $z$:
Let
$$
Q = int_S x^2+y^2+z^2 ~ dA
$$, where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.
Substituting $u = z-1; u+1 = z; du = dz$, we get
$$
Q = int_T x^2+y^2+(u+1)^2 ~ dA
$$, where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
$$
Q = int_U x^2+y^2+(z+1)^2 ~ dA,
$$
where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.
Expanding the integrand, we get
begin{align}
Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
&= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 2 ~dA+ int_U 2z~dA\
&= 4pi+ 2int_U z~dA
end{align}
The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.
Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
$$
(x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
$$
is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.
So the overall value of your integral is $6pi$.
answered Nov 27 at 13:42
John Hughes
62.3k24090
add a comment |
Hoping that by "upper" hemisphere, you mean "with respect to $z$:
Let
$$
Q = int_S x^2+y^2+z^2 ~ dA
$$, where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.
Substituting $u = z-1; u+1 = z; du = dz$, we get
$$
Q = int_T x^2+y^2+(u+1)^2 ~ dA
$$, where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
$$
Q = int_U x^2+y^2+(z+1)^2 ~ dA,
$$
where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.
Expanding the integrand, we get
begin{align}
Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
&= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 2 ~dA+ int_U 2z~dA\
&= 4pi+ 2int_U z~dA
end{align}
The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.
Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
$$
(x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
$$
is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.
So the overall value of your integral is $6pi$.
answered Nov 27 at 13:42
John Hughes
62.3k24090
add a comment |
Hoping that by "upper" hemisphere, you mean "with respect to $z$:
Let
$$
Q = int_S x^2+y^2+z^2 ~ dA
$$, where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.
Substituting $u = z-1; u+1 = z; du = dz$, we get
$$
Q = int_T x^2+y^2+(u+1)^2 ~ dA
$$, where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
$$
Q = int_U x^2+y^2+(z+1)^2 ~ dA,
$$
where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.
Expanding the integrand, we get
begin{align}
Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
&= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 2 ~dA+ int_U 2z~dA\
&= 4pi+ 2int_U z~dA
end{align}
The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.
Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
$$
(x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
$$
is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.
So the overall value of your integral is $6pi$.
answered Nov 27 at 13:42
John Hughes
62.3k24090
Hoping that by "upper" hemisphere, you mean "with respect to $z$:
Let
$$
Q = int_S x^2+y^2+z^2 ~ dA
$$, where $S$ is the upper hemisphere of the sphere $x^2+y^2+(z-1)^2=1$.
Substituting $u = z-1; u+1 = z; du = dz$, we get
$$
Q = int_T x^2+y^2+(u+1)^2 ~ dA
$$, where $T$ is the upper hemisphere of the sphere $x^2+y^2+u^2=1$, which I'll rewrite as
$$
Q = int_U x^2+y^2+(z+1)^2 ~ dA,
$$
where $U$ is the upper hemisphere of the sphere $x^2+y^2+z^2=1$.
Expanding the integrand, we get
begin{align}
Q &= int_U x^2+y^2+z^2 + 2z + 1 ~ dA \
&= int_U x^2+y^2+z^2 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 1 ~dA+ int_U 2z~dA + int_U 1 ~ dA \
&= int_U 2 ~dA+ int_U 2z~dA\
&= 4pi+ 2int_U z~dA
end{align}
The latter integral is the average height of a point on the hemisphere characterized by $z ge 0$, muktiplied by the area ($2pi$) of that hemisphere.
Because horizontal radial projection from the $z$-axis to a cylinder surrounding the sphere, i.e., the map
$$
(x, y, z) mapsto (frac{x}{sqrt{x^2 + y^2}}, frac{y}{sqrt{x^2 + y^2}}, z)
$$
is area-preserving (which requires a little proof, but was mentioned by Boy in his dissertation somewhere around 1895-1905, so it's a pretty well-established result!), this is the same as the average-height-times-area of a point on a cylinder of radius $1$ and height $1$, i.e., $frac{1}{2} cdot 2pi = pi$.
So the overall value of your integral is $6pi$.
answered Nov 27 at 13:42
John Hughes
62.3k24090
answered Nov 27 at 13:42
John Hughes
62.3k24090
answered Nov 27 at 13:42
John Hughes
62.3k24090
answered Nov 27 at 13:42
John Hughes
62.3k24090
62.3k24090
add a comment |
add a comment |
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