$\Sigma$ 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 $D_{x y}=\left\{(x, y) \mid x^{2}+y^{2} \leq 4\right\}$.
$$ \iint_{\Sigma} \sqrt{4-x^{2}-4 z^{2}} \mathrm{d} \mathrm{x} \mathrm{dy}=\iint_{\Sigma}|y| \mathrm{d} \mathrm{x} \mathrm{dy}=\iint_{x^{2}+y^{2} \leq 4}|y| \mathrm{d} \mathrm{x} \mathrm{dy}\\ = 2 \int_{0}^{\pi} d \theta \int_{0}^{2} r^{2} \sin \theta d r =\frac{32}{3} $$
There are the following conversion formulas in the textbook:
$$\iint_{\Sigma} R(x, y, z) d x d y=\pm \iint_{D_{x y}} R[x, y, z(x, y)] d x d y$$
$$\iint_{\Sigma} P(x, y, z) d y d z=\pm \iint_{D_{y z}} P[x(y, z), y, z] d y d z$$
$$\iint_{\Sigma} Q(x, y, z) d x d z=\pm \iint_{D_{x z}} Q[x, y(x, z), z] d x d z$$
$$\begin{array}{l} \iint_{x} P(x, y, z) \mathrm{d} y \mathrm{d} z+Q(x, y, z) \mathrm{d} z \mathrm{d} x+R(x, y, z) \mathrm{d} x \mathrm{d} y \\ =\iint[P(x, y, z) \cos \alpha+Q(x, y, z) \cos \beta+R(x, y, z) \cos \gamma] \mathrm{d} S \\ =\pm \iint_{D}\left[P(x(u, v), y(u, v), z(u, v)) \frac{\partial(y, z)}{\partial(u, v)}+Q(x(u, v), y(u, v)\right. \\ \left.z(u, v)) \frac{\partial(z, x)}{\partial(u, v)}+R(x(u, v), y(u, v), z(u, v)) \frac{\partial(x, y)}{\partial(u, v)}\right] \mathrm{d} u \mathrm{d} v\\ \overset{In particular, when z=z(x,y)}{\Longrightarrow} \pm \iint_{D}\left [-P(x, y, z(x, y)) \frac{\partial z(x,y)}{\partial(x)}-Q(x, y\right. \\ \left.z(x,y)) \frac{\partial z(x,y)}{\partial y}+R(x, y, z(x,y)) \right] \mathrm{d} x \mathrm{d} y \end{array}$$
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 ∑*)
$$\begin{array}{c} \vec{n}=(\cos \alpha, \cos \beta, \cos \gamma)=\frac{1}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}}\left(-z_{x},-z_{y}, 1\right) \\ \therefore \cos \gamma=\frac{1}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}} \\ \frac{d x d y}{d s}=\cos \gamma=\frac{1}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}} \\ \frac{d y d z}{d s}=\cos\alpha=-\frac{z_{x}}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}} \\ \frac{d z d x}{d s}=\cos \beta=-\frac{z_{y}}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}}\\ \therefore \iint_{\Sigma} R(x, y, z) d x d y=\iint_{\Sigma} R(x, y, z) \cos \gamma d S \\ =\iint_{\Sigma} R[x, y, z(x, y)] \cdot \frac{1}{\sqrt{1+z_{x}^{2}+z y^{2}}} d S \\ =\iint_{D_{x y}} R[x, y, z(x, y)] \cdot \frac{1}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}} \cdot \sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y \\ =\iint_{D_{xy}} R[x, y, z(x, y)] d x d y \end{array}$$
$$\begin{array}{c} \iint_{\Sigma} Q(x, y, z) d y d z=\iint_{\Sigma} Q(x, y, z) \cos \beta d S \\ =\iint_{\Sigma} Q[x, y, z(x, y)] \cdot \frac{-z_{x}}{\sqrt{1+z_{x}^{2}+z y^{2}}} d S \\ =\iint_{D_{xy}} Q[x, y, z(x, y)] \cdot \frac{-z_{x}}{\sqrt{1+z_{x}^{2}+z_{y}^{2}}} \cdot \sqrt{1+z_{x}^{2}+z_{y}^{2}} d x d y \\ =\iint_{D_{xy}} -Q[x, y, z(x, y)]z_{x} d x d y \end{array}$$
Disk[{0,0},2]and multiply by the area element - I will update my answer. $\endgroup$Integrate[Sqrt[4 - x^2 - 4z^2], {x, y, z} ∈ reg]$\endgroup$DoubleContourIntegralis because you're integrating a vector field that's pointing directly up:DoubleContourIntegral[{0, 0, Sqrt[4 - x^2 - 4 z^2]},...and you have a dot product with the normal that measures the flux of the field. So in that sense it's correct that it's 32/3, however, the other surface integral that you're having problems with is scalar - it accumulates a scalar field over infinitesimal surface patches, and the dot product doesn't make sense in this context. $\endgroup$DoubleContourIntegral[{0, 0, 1}, {({x, y, z} -> {Sqrt[1 - v^2], Sqrt[1 - u^2], z}), {u, -1, 1}, {v, -Sqrt[1 - u^2], Sqrt[1 - u^2]}}]we get zero. However, a scalar surface integral of 1 would give the true surface area (z Pi). $\endgroup$