Probational Polar Positions

Calculus Level 3

$\large \int_0^{\frac {\sqrt 3}{2} } \int_0^1 \frac {1}{ \sqrt{1-x^2} \sqrt{1-y^2} } \ dx \ dy = \ ?$

Give your answer to two decimal places.

5 solutions

Jaber Al-arbash
Nov 16, 2014

integral 1/( sqrt(1- x^2)) = sin^-1x

integral 1/(sqrt(1-y^2)) = sin^-1y

By plugging in the numbers we get:

(pi/2)(pi/3) = pi^2/6 = 1.64

Anna Anant
Nov 17, 2014

1/( sqrt(1- x^2)) = sin^-1x Integral 1/(sqrt(1-y^2)) = sin^-1y By putting the numbers we get: =[(pi/2)(pi/3)] = [(pi)^2 / 6] = 1.64

Rohan Shah
Nov 17, 2014

∫1/√(1-x²) = -cos⁻¹(x)

∫1/√(1-y²) = -cos⁻¹(y)

[{-cos⁻¹(1)} - {-cos⁻¹(0)} = Π/2

[{-cos⁻¹(√3/2)} - {-cos⁻¹(0)} = Π/3

Π/2 x Π/3 = Π²/6 = 1.64

Suvaditya Sur
Nov 19, 2014

integrate 1st by wrt dx and then put limits and den integrate wrt dy

Goli Haveesh
Nov 17, 2014

Integrate f(x) keeping f(y) constant and then integrate f(y) or vice versa

You are just giving a definition of integrals.

Joel Tan - 6 years, 6 months ago