# Evaluate the integral: integral_0^1 integral_0^{x square root 3} ( {1} / {(1 + x^2 + y^2)^2} ) dy...

## Question:

Evaluate the integral:

{eq}\displaystyle \int_0^1 \int_0^{x \sqrt 3} \bigg( \frac{1} {(1 + x^2 + y^2)^2} \bigg)\ dy\ dx {/eq}.

## Evaluating Double Integral Using Polar Coordinates

Given a definite double integral over a closed and bounded domain in the shape of a right triangle, we use polar coordinates to evaluate the integral. We convert Cartesian coordinates to polar coordinates using traditional techniques from Calculus and then evaluate the resulting definite integrals using the Fundamental Theorem of Calculus which involves the use of anti-derivatives.

The region of integration is a right triangle with coordinates {eq}(0, 0), \; (1, 0) \; {\rm and} \; (1, \sqrt {3}). {/eq}

The angle the hypotenuse of this right triangle makes with the x-axis is {eq}\pi/3 {/eq} radians.

Next we use the following conversion formula from Cartesian to polar form:

{eq}x=r\cos \theta, \; y = r \sin \theta, \; x^2+y^2=r^2, \; dy \; dx = r \; dr \; d\theta. {/eq}

Further, using right triangle Trigonometry in our region of integration we get that our double integral equals the following:

{eq}\displaystyle I=\int_0^{\pi/3} \int_0^{1/\cos \theta} \frac {r \; dr \; d\theta}{(1+r^2)^2} \qquad (1) {/eq}

Use the u-substitution in the inner r-integral {eq}u=1+r^2 \implies du = 2r \; dr {/eq} with new limits of integration

{eq}\displaystyle r=0 \implies u=1+0^2=1 \; {\rm and} \; r=\frac {1}{\cos \theta} \implies u = \frac{1+\cos^2 \theta}{\cos^2 \theta} {/eq} leads to the integral in (1) equaling

{eq}\displaystyle I=\int_0^{\pi/3} \left( \int_1^{\frac {1+\cos^2 \theta}{\cos^2 \theta}} \frac {du}{2u^2} \right) \; d \theta = \frac 12 \int_0^{\pi/3} \left[ -\frac 1u \right]_1^{\frac {1+\cos^2 \theta}{\cos^2 \theta}} \; d\theta = \frac 12 \int_0^{\pi/3} \left( 1- \frac {\cos^2 \theta}{1+\cos^2 \theta} \right) \; d\theta = \frac 12 \int_0^{\pi/3} \frac {d\theta}{1+\cos^2 \theta} \qquad (2) {/eq}

Using technology such as a graphing calculator gives us that the value of the definite integral in (2) equals 0.31.