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I want to get the area of the set $\{x^2 + y\, | \,x^2 + y^2 < 2\}$, meaning I want to get the area of the surface defined by

$\qquad f(x,y) = x^2 + y$

with the domain given by $x^2 + y^2 < 2$.

I tried using ImplicitRegion, and I can even plot it with 

Plot3D[x^2 + y, {x, y} ∈ ImplicitRegion[x^2 + y^2 < 2, {x, y}]]

but I don't know how to measure the area of the surface.

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    $\begingroup$ It is unclear what you're asking. What does it mean the "area" of $x^2 + y$? And why is your domain $x^2 + y^2 < 2$ but your Plot3D have an ImplicitRegion $x^2 + y^2 < 9$? Please fix your question! $\endgroup$
    – David G. Stork
    Dec 14 '17 at 21:51
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    $\begingroup$ The Plot3D also plots a different function (2x+y) instead of (x^2+y). $\endgroup$
    – Cassini
    Dec 14 '17 at 21:58
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I believe the question is about the area of a surface. If that's the case, the answer is as follows:

surface = ImplicitRegion[x^2 + y == z && x^2 + y^2 < 2, {x, y, z}]

This is a 2D region embedded in 3D space:

RegionDimension[surface]
RegionEmbeddingDimension[surface]
(*2*)
(*3*)

The area is found by using RegionMeasure:

RegionMeasure[surface]
N[%, 10]
(*2/3 Sqrt[2] (3 EllipticE[-4] + 5 EllipticK[-4])*)
(*12.21203137*)
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  • $\begingroup$ This is exactly what I wanted, thanks a lot! $\endgroup$
    – Csharphobbydev
    Dec 15 '17 at 15:21
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Integrate[x*x + y, {x, y} ∈ Disk[{0, 0}, Sqrt [2]]]

which returns $\pi$.

However, if it is school work, the "manual" way to do that is a change of variable like this one: $(x,y)=(r\cos \theta,r\sin \theta)$

After Jacobian calculation you get: 

Integrate[r*(r*r*Cos[θ]^2 + r*Sin[θ]), {r, 0, Sqrt[2]}, {θ, 0, 2 π}]

Which (fortunately) also gives $\pi$

Update: Now if you want area of the surface, you can use this formula: $$ \int_\Omega \| \left( \begin{array}{c} 1 \\ 0 \\ \partial_x f \end{array} \right)\times \left( \begin{array}{c} 0 \\ 1 \\ \partial_y f \end{array} \right) \| dxdy $$ with $\Omega$ your centered disk of radius $\sqrt{2}$ and $f:(x,y)\rightarrow x^2+y$. You get, after change of variable $$ \int_0^\sqrt{2}\int_0^{2\pi} r \sqrt{4 r^2 \cos ^2(\theta )+2}\ drd\theta $$ which unfortunately involves Elliptic functions...

Under Mathematica:

f[x_, y_] := x*x + y

vx = D[{x, y, f[x, y]}, x];
vy = D[{x, y, f[x, y]}, y];

dareaCartesian = 
 Simplify[Norm[Cross[vx, vy], 2], Assumptions -> {x ∈ Reals}]

dareaPolar = 
 r*Simplify[dareaCartesian /. x -> r*Cos[θ] /. y -> r*Sin[θ] ]

area = Integrate[dareaPolar, {r, 0, Sqrt[2]}, {θ, 0, 2*Pi}] 

N[area]

which prints: $$ \sqrt{4 x^2+2} $$ $$ r \sqrt{4 r^2 \cos ^2(\theta )+2} $$ $$ \frac{2}{3} \sqrt{2} (5 EllipticK(-4)+3 EllipticE(-4)) $$ $$ 12.212 $$

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    Integrate[(x^2 + y) Boole[x^2 + y^2 < 2], {x, -Infinity, 
   Infinity}, {y, -Infinity, Infinity}]

Pi

(* If you want a plot, do this: *)

 Plot3D[(x^2 + y) Boole[x^2 + y^2 < 2], {x, -Sqrt[2], 
  Sqrt[2]}, {y, -Sqrt[2], Sqrt[2]}, 
 RegionFunction -> (#1^2 + #2^2 < 2 &), Filling -> Axis, 
 FillingStyle -> Directive[Gray, Opacity[0.2]]]
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