Double Integral

[JJ007]

Use polar coordinates to evaluate \(\displaystyle \int_{-1}^{1}\int_0^\sqrt{1-x^2}\)\(\displaystyle {e^{x^2+y^2}}dydx\)

\(\displaystyle \int_{-1}^{1}\int_0^{1-cos(theta)}\)\(\displaystyle e^r\ dr\ d(theta)?\)

 

[galactus]

Don't forget that extra r in polar.

\(\displaystyle \frac{1}{2}\int_{0}^{2\pi}\int_{0}^{1}re^{r^{2}}drd{\theta}\)

 

[Deleted member 4993]

Don't forget that extra r in polar.<<< That makes the integration easy

\(\displaystyle \frac{1}{2}\int_{0}^{2\pi}\int_{0}^{1}re^{r^{2}}drd{\theta}\)

 

[Deleted member 4993]

Use polar coordinates to evaluate \(\displaystyle \int_{-1}^{1}\int_0^\sqrt{1-x^2}\)\(\displaystyle {e^{x^2+y^2}}dydx\)

\(\displaystyle \int_{-1}^{1}\int_0^{1-cos(theta)}\)\(\displaystyle e^r\ dr\ d(theta)?\)

How did you get:

\(\displaystyle x^2 \ \ + \ \ y^2 \ \ = \ \ r^1\)

and

\(\displaystyle \sqrt{1-x^2} \ \ = \ \ 1 \ \ - \ \ cos(\theta)\)

These "mistakes" will lead you to wrong path - look at galactus's response for correct limits and substitutions.

 

[JJ007]

Use polar coordinates to evaluate \(\displaystyle \int_{-1}^{1}\int_0^\sqrt{1-x^2}\)\(\displaystyle {e^{x^2+y^2}}dydx\)

\(\displaystyle \int_{-1}^{1}\int_0^{1-cos(theta)}\)\(\displaystyle e^r\ dr\ d(theta)?\)

How did you get:

\(\displaystyle x^2 \ \ + \ \ y^2 \ \ = \ \ r^1\)

and

\(\displaystyle \sqrt{1-x^2} \ \ = \ \ 1 \ \ - \ \ cos(\theta)\)

These "mistakes" will lead you to wrong path - look at galactus's response for correct limits and substitutions.

My textbook had a different method for polar coordinates when switching the order of integration. I got mixed up.