- #1
Hiero
- 322
- 68
Homework Statement
(Scroll to bottom for the true question)
Suppose we are to find the integral from -∞ to +∞ of (let’s just say) e-x2dx
Homework Equations
∫∫f(x)g(y)dxdy = (∫f(x)dx)(∫g(y)dy)
The Attempt at a Solution
We can square the integral we want to solve for then use my relevant equation (in reverse) to write the answer to the integral as the square root of this double integral:
∫∫e-(x2+y2)dxdy
Where the x and y boundaries are ±∞
Now we transform into polar coordinates so that the integral becomes:
∫∫e-r2(rdrdθ)
Now we could make the (inner) r integral run from 0 to ∞ and the (outer) θ integral run from 0 to 2π (since these boundaries cover the whole plane) which would give the famous answer.Now my question is about the boundaries of the polar integral... why can’t we have made r run from -∞ to +∞ and θ run from 0 to π? This also seems to cover the plane, but doing this gives an answer of zero. What is the reason this is wrong?
Thanks.
Well actually I just realized the “r” in rdrdθ comes from the absolute value of the ”Jacobian determinant” so it should actually be |r| dr dθ which does give the correct answer.
I’m going to post this anyway since I already typed it out. Maybe someone has more insight to share.