Question

double integral sec(x^2 +y^2) dx dy . Use polar coordinates

Answer 1

Also, x^2+y^2=1

Answer 2

0 to 1 ^

Answer 3

\[\iint_D \sec(x^2+y^2)\ dx\ dy\]
What's the region of integration?

Answer 4

x^2 +y^2=<1

Answer 5

Okay, so in polar coordinates, that region can be written like this:
\[D=\{(r,\theta)|0\leq r\leq1, 0\leq \theta \leq 2\pi\}\]
Then rewrite the double integral like this:
\[\int_0^{2\pi} \int_0^1 \sec(r^2)\ r\ dr\ d\theta\]
I know you can evaluate that. :)

Answer 6

thanx. This one is tedious