# Double Integrals

• Mar 27th 2009, 10:07 PM
ishanj07
Double Integrals
Can anybody help me with this?
See Attachment

Thanks,
Ishan
• Mar 27th 2009, 10:43 PM
matheagle
Use $0\le x^2+y^2\le 1$ in your exponent and integrate over the region.

So $1=e^0\le e^{4(x^2+y^2)}\le e^4$

and $\pi=\int\int_D DA \le \int\int_D e^{4(x^2+y^2)}DA\le e^4\int\int_D DA=e^4\pi$