# Calculate a double integrate over an implicit region using W|A?

GROUPS:
 Hello! How do I double integrate over an implicit region, like an elipsis, using the wolframAlpha?
Answer
10 days ago
3 Replies
 Marco Thiel 1 Vote Hi Thiago,I could suggest to use the free (!) tier of the Development Cloud instead:https://www.wolframcloud.comIt is more flexible and easier to make concise requests if you know a bit of the Wolfram Language. It looks like this:So the code you would use is exactly the code in your related thread.Cheers,Marco
Answer
10 days ago
 Thanks, Marco. But, I'm looking for the "Step-by-Step" solution thats is built-in on Wolfram, thats the reason I mentioned it
Answer
10 days ago
 Marco Thiel 1 Vote Hi, I do not think that you will get an automatic step-by-step solution for this. But this looks like a textbook example, which is easy to solve. You still integrate over a disk - a shifted unit disk: ContourPlot[x^2 + y^2 - 2 y == 0, {x, -2, 3}, {y, -2, 3}] So if you introduce the new variable z->y-1, you obtain a unit disk: Simplify[x^2 + y^2 - 2 y == 0 /. y -> z + 1] (*x^2 + z^2 == 1*) So the new integral is Integrate[1, {x, z} \[Element] Disk[]] In polar coordinates this is: Integrate[r, {\[Phi], 0, 2 Pi}, {r, 0, 1}] which evaluates to Pi. So even if Mathematica doesn't show you the step-by-step solution it is really easy to perform a step by step calculation. Cheers, Marco
Answer
10 days ago