Mathematica Asked on October 3, 2021
$Sigma$ is the upper side of the surface $x^{2}+y^{2}+4 z^{2}=4(z geq 0)$, now we need to calculate the value of $iint_{Sigma} sqrt{4-x^{2}-4 z^{2}} d x d y$ (second-kind surface integral).
When I use the following code to calculate directly, the result is incorrect :
reg = Region[
ImplicitRegion[x^2 + y^2 + 4 z^2 == 4 && z >= 0, {x, y, z}]]
Integrate[Sqrt[4 - x^2 - 4z^2], {x, y, z} ∈ reg]
When I refer to the code in this post for calculation, I can’t get the correct result(the answer should be $frac{32} {3}$). In addition, I want to know why this method can’t get the right results, and what’s wrong with my thinking.
region = Region[
ImplicitRegion[x^2 + y^2 + 4 z^2 == 4 && z >= 0, {x, y, z}]];
Integrate[#, {x, y, z} ∈ region] & /@ ({0, 0, Sqrt[
4 - x^2 - 4 z^2]}.Normalize[{x, y, 4 z}])
But when I use the code of this post, I can get the result of $frac{32}{3}$.
DoubleContourIntegral[field_?VectorQ,
surface : {changeOfVars : ({x_, y_, z_} ->
param : {xuv_, yuv_, zuv_}), {u_, u1_, u2_}, {v_, v1_, v2_}}] :=
Integrate[
Dot[field /. Thread[changeOfVars],
Cross[D[param, u], D[param, v]]], {u, u1, u2}, {v, v1,
v2}];
DoubleContourIntegral[{0, 0,
Sqrt[4 - x^2 -
4 z^2]}, {({x, y, z} -> {u, v, Sqrt[4 - u^2 - v^2]/2}), {u, -2,
2}, {v, -Sqrt[4 - u^2], Sqrt[4 - u^2]}}]
What should I do?
Analysis process for reference:
The projection area of surface $Sigma$ on xOy plane is $D_{x y}=left{(x, y) mid x^{2}+y^{2} leq 4right}$.
$$
iint_{Sigma} sqrt{4-x^{2}-4 z^{2}} mathrm{d} mathrm{x} mathrm{dy}=iint_{Sigma}|y| mathrm{d} mathrm{x} mathrm{dy}=iint_{x^{2}+y^{2} leq 4}|y| mathrm{d} mathrm{x} mathrm{dy}
= 2 int_{0}^{pi} d theta int_{0}^{2} r^{2} sin theta d r
=frac{32}{3}
$$
There are the following conversion formulas in the textbook:
$$iint_{Sigma} R(x, y, z) d x d y=pm iint_{D_{x y}} R[x, y, z(x, y)] d x d y$$
$$iint_{Sigma} P(x, y, z) d y d z=pm iint_{D_{y z}} P[x(y, z), y, z] d y d z$$
$$iint_{Sigma} Q(x, y, z) d x d z=pm iint_{D_{x z}} Q[x, y(x, z), z] d x d z$$
$$begin{array}{l}
iint_{x} P(x, y, z) mathrm{d} y mathrm{d} z+Q(x, y, z) mathrm{d} z mathrm{d} x+R(x, y, z) mathrm{d} x mathrm{d} y
=iint[P(x, y, z) cos alpha+Q(x, y, z) cos beta+R(x, y, z) cos gamma] mathrm{d} S
=pm iint_{D}left[P(x(u, v), y(u, v), z(u, v)) frac{partial(y, z)}{partial(u, v)}+Q(x(u, v), y(u, v)right.
left.z(u, v)) frac{partial(z, x)}{partial(u, v)}+R(x(u, v), y(u, v), z(u, v)) frac{partial(x, y)}{partial(u, v)}right] mathrm{d} u mathrm{d} v
overset{In particular, when z=z(x,y)}{Longrightarrow} pm iint_{D}left [-P(x, y, z(x, y)) frac{partial z(x,y)}{partial(x)}-Q(x, yright.
left.z(x,y)) frac{partial z(x,y)}{partial y}+R(x, y, z(x,y)) right] mathrm{d} x mathrm{d} y
end{array}$$
In particular, when $z=z(x,y)$:
DForm[f_] :=
f /. {Derivative[inds__][g_][vars__] :> Which[Length[{vars}] >= 2,
Apply[Defer[D[g[vars], ##]] &,
Transpose[{{vars}, {inds}}] /. {{v_, 1} :> {v}, {v_, 0} :>
Nothing}], Length[{vars}] == 1,
Apply[Defer[Dt[g[vars], #]] &,
Transpose[{{vars, inds}}] /. {{v_, 1} :> {v}}]]} //
TraditionalForm ;
jacobian[u_, v_, {x_, y_}] := Det[( {
{D[u, x], D[u, y]},
{D[v, x], D[v, y]}
} )] // DForm
(*In particular,when z=z(x,y)*)
jacobian[y, z[x, y], {x, y}]
jacobian[z[x, y], x, {x, y}]
jacobian[x, y, {x, y}](*Inner side of surface ∑*)
-jacobian[x, y, {x, y}](*Outside of surface ∑*)
$$begin{array}{c}
vec{n}=(cos alpha, cos beta, cos gamma)=frac{1}{sqrt{1+z_{x}^{2}+z_{y}^{2}}}left(-z_{x},-z_{y}, 1right)
therefore cos gamma=frac{1}{sqrt{1+z_{x}^{2}+z_{y}^{2}}}
frac{d x d y}{d s}=cos gamma=frac{1}{sqrt{1+z_{x}^{2}+z_{y}^{2}}}
frac{d y d z}{d s}=cosalpha=-frac{z_{x}}{sqrt{1+z_{x}^{2}+z_{y}^{2}}}
frac{d z d x}{d s}=cos beta=-frac{z_{y}}{sqrt{1+z_{x}^{2}+z_{y}^{2}}}
therefore iint_{Sigma} R(x, y, z) d x d y=iint_{Sigma} R(x, y, z) cos gamma d S
=iint_{Sigma} R[x, y, z(x, y)] cdot frac{1}{sqrt{1+z_{x}^{2}+z y^{2}}} d S
=iint_{D_{x y}} R[x, y, z(x, y)] cdot frac{1}{sqrt{1+z_{x}^{2}+z_{y}^{2}}} cdot sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y
=iint_{D_{xy}} R[x, y, z(x, y)] d x d y
end{array}$$
$$begin{array}{c}
iint_{Sigma} Q(x, y, z) d y d z=iint_{Sigma} Q(x, y, z) cos beta d S
=iint_{Sigma} Q[x, y, z(x, y)] cdot frac{-z_{x}}{sqrt{1+z_{x}^{2}+z y^{2}}} d S
=iint_{D_{xy}} Q[x, y, z(x, y)] cdot frac{-z_{x}}{sqrt{1+z_{x}^{2}+z_{y}^{2}}} cdot sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y
=iint_{D_{xy}} -Q[x, y, z(x, y)]z_{x} d x d y
end{array}$$
I've been back and forth trying to work out why Mathematica was getting the surface integral 'wrong'. But Mathematica can do surface integrals properly. See this answer. It turns out your original calculation with the region is correct after all:
reg = ImplicitRegion[x^2 + y^2 + 4 z^2 == 4 && z >= 0, {x, y, z}];
Integrate[Sqrt[4 - x^2 - 4z^2], {x, y, z} ∈ reg]
(* result: 32/9 (2 EllipticE[3/4] + EllipticK[3/4]) *)
N[%]
(* 16.2796 *)
Note: My answer before of 32/3 was wrong because an incorrect usage of nested With and the derivative was giving the wrong surface element of 1.
All that remains is to show that your integrals are incorrect and that the surface integrals are a lot more complicated:
$$ iint_{Sigma} sqrt{4-x^{2}-4 z^{2}} mathrm{d}Sigma=iint_{Sigma}|y| mathrm{d} Sigma $$
We substitute $x=rcos(theta)$, $y=rsin(theta)$ and $z=frac{1}{2} sqrt{4-x^2-y^2}=frac{1}{2} sqrt{4-r^2}$. The element of area for cylindrical coordinates is $r mathrm{d}r mathrm{d}theta$ (see $S_z$ here). This means that:
$$ mathrm{d}Sigma=left|frac{partial{Sigma}}{partial{r}}timesfrac{partial{Sigma}}{partial{theta}}right|cdot r mathrm{d}rmathrm{d}theta=r sqrt{frac{1}{4} left| frac{r}{sqrt{4-r^2}}right| ^2+1} $$
$$ iint_{Sigma} sqrt{4-x^{2}-4 z^{2}} mathrm{d}Sigma=int_{0}^{2}int_{0}^{2pi}|rsin(theta)|cdot r sqrt{frac{1}{4} left| frac{r}{sqrt{4-r^2}}right| ^2+1} mathrm{d}theta mathrm{d}r $$
The result of this integral is quite messy, involving elliptical functions.
I will show how to calculate it in both cartesian and cylindrical:
Cartesian:
With[{z2 = (4 - x^2 - y^2)/4},
With[{z = Sqrt[z2]},
Integrate[
Sqrt[4 - x^2 - 4 z2] Sqrt[1 + D[z, x]^2 + D[z, y]^2], {x,
y} [Element] Disk[{0, 0}, 2]]
]
] // FullSimplify
(* 8/27 (-24 - 24 I Sqrt[2] 3^(1/4) EllipticE[1/2 - 7/(8 Sqrt[3])] -
12 (-2 + Sqrt[3]) EllipticE[-8 (12 + 7 Sqrt[3])] +
12 (-2 + Sqrt[3]) EllipticE[
I ArcCsch[3^(1/4) + 3^(3/4)], -8 (12 + 7 Sqrt[3])] -
60 (2 + Sqrt[3]) EllipticF[
I ArcCsch[3^(1/4) + 3^(3/4)], -8 (12 + 7 Sqrt[3])] +
I Sqrt[6 (168 + 97 Sqrt[3])] EllipticK[1/2 - 7/(8 Sqrt[3])] +
60 I (2 + Sqrt[3]) EllipticK[97 + 56 Sqrt[3]]) *)
N[%]
(* 16.2796 *)
Cylindrical:
z = Sqrt[4 - r^2]/2;
sz[r_, [Theta]_] := {r, [Theta], z}
el = Norm[Cross[D[sz[r, [Theta]], r], D[sz[r, [Theta]], [Theta]]]];
Integrate[Abs[r Sin[[Theta]]]*el*r, {[Theta], 0, 2 [Pi]}, {r, 0, 2}]
(* 8/27 (-24 - 24 I Sqrt[2] 3^(1/4) EllipticE[1/2 - 7/(8 Sqrt[3])] +
24 EllipticE[-8 (12 + 7 Sqrt[3])] -
12 Sqrt[3] EllipticE[-8 (12 + 7 Sqrt[3])] -
24 EllipticE[I ArcCsch[3^(1/4) + 3^(3/4)], -8 (12 + 7 Sqrt[3])] +
12 Sqrt[3]
EllipticE[I ArcCsch[3^(1/4) + 3^(3/4)], -8 (12 + 7 Sqrt[3])] -
120 EllipticF[I ArcCsch[3^(1/4) + 3^(3/4)], -8 (12 + 7 Sqrt[3])] -
60 Sqrt[3]
EllipticF[I ArcCsch[3^(1/4) + 3^(3/4)], -8 (12 + 7 Sqrt[3])] +
I Sqrt[6 (168 + 97 Sqrt[3])] EllipticK[1/2 - 7/(8 Sqrt[3])] +
120 I EllipticK[97 + 56 Sqrt[3]] +
60 I Sqrt[3] EllipticK[97 + 56 Sqrt[3]]) *)
N[%]//Chop
(* 16.2796 *)
Both of these numerical results agree with the result from integrating over the ImplicitRegion except the expressions weren't as simple. Presumably there are some different coordinate transformations going on internally which leads to the nicer expression 32/9 (2 EllipticE[3/4] + EllipticK[3/4]).
Correct answer by flinty on October 3, 2021
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