r = e^{−θ/10}, π/2 ≤ θ ≤ π

r^{2} = e^{-theta}^{/5}

1/2[Integral e^{-theta}^{/5} d theta between pi/2 to pi]

1/2[( -5 e^{-}^{theta/5}) between pi/2 to pi]

= -5/2 ( e^{-}^{pi/5}- e^{-}^{pi/10})=5/2(e^{-}^{pi/10}- e^{-}^{pi/5})

= 5/2(0.730403-0.533488)=

=5/2*0.196915 =0.984573/2= 0.492286

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up