1 point dr dy e R2 A Using polar coordinates evaluate the im

1 point) dr dy. e- R2 A. Using polar coordinates, evaluate the improper integral B. Use part A to evaluate the improper integral e-7* dx.

Solution

(A)
e^(-7(x² + y²)) dx dy

over the entire real plane R².
It can be re- written as -

e^(-7(x² + y²)) = e^(-7x²) e^(-7y²)

That is, the problem is the product of two functions, one of which doesn\'t mention y and the other of which doesn\'t mention x. This means that the double integral is separable:

e^(-7(x² + y²)) dx dy
= ( e^(-7x²) dx) × ( e^(-7y²) dy)
= ( e^(-7x²) dx)²

where, again, all integrals are from - to .

Alternatively, we can perform a change of coordinates from Cartesian to polar, noting that by Pythagoras\' theorem, x² + y² = r²:

e^(-7(x² + y²)) dx dy
= e^(-7r²) r dr d

where the integral of r is from 0 to and the integral of is from 0 to 2.

Again, this is a separable double integral. Using the substitution:
u = -7r²
du/dr = -20r
-du/20 = r dr

we have:

e^(-7r²) r dr d
= ( e^(-7r²) r dr) × ( d)
= 2 e^(-7r²) r dr
= /7 e^u du

where the bounds on the integral are now from u=- to u=0. The integral is trivial, and we find:

/7 e^u du = /7

So , we have:

e^(-7(x² + y²)) dx dy
= ( e^(-7x²) dx)²
= /7

So we have the answer for the double integral.

(B) For the single integral, we take the positive square root to find:

e^(-7x²) dx = (/7)