Integration of an area given by an inequality does not work for me?

Questions
,
1

Hello,
I had to find the area of M with M :={(x,y) ∈ IR² | x²+(e^x + y)^2 <= 1 }. I did this on a paper and got v(M) = Pi. I wanted to verify my solution by using Mathematica.
I was able to plot the area with

RegionPlot[{x^2 + (Exp[x] + y)^2 <= 1}, {x, -2, 2}, {y, -4, 2}]

I also was able to find the limits of integration with

Reduce[x^2 + (Exp[x] + y)^2 <= 1]

But unfortunately, I am unable to directly calculate the area. Commands like

Area[{x^2 + (Exp[x] + y)^2 <= 1}, {x, -2, 2}, {y, -4, 2}]

just gives me the result 0.

Integrate[{x^2 + (Exp[x] + y)^2 <= 1}, {x, -2, 2}, {y, -4, 2}]

just prints the Integral but does not evaluate it.

Can anyone help me please?

2

Define an implicit region for the constraints you imposed

r= 
 ImplicitRegion[
  x^2 + (Exp[x] + y)^2 <= 1 && -2 <= x <= 2 && -4 <= y <= 2, {x, y}]

Then just compute the area

Area[r]

and obtain your expected result.

3

This worked for me. And I got PI/2 as an answer.

Integrate[{(E^x + y)^2 + x^2}*Boole[(E^x + y)^2 + x^2 <= 1], {x, -2, 
  2}, {y, -4, 2}]
4

This is incorrect.
You area element is dx dy if x and y are within the specified area. So, If you want to take this approach you need use the following

Integrate[Boole[(E^x + y)^2 + x^2 <= 1], {x, -2, 2}, {y, -4, 2}]

which returns Pi and agrees with the result of the OP and the area of the specified region (see above)

5

Of course, my bad. I copied the code and applied the functional value within the region.