# Calculus: Change the Cartesian Integral into an Equivalent Polar Integral

**Change the Cartesian Integral into an equivalent polar integral then evaluate the polar integral.**

Given Cartesian Integral: ∫0-1∫0-sqrt(1-y^2) (x^2 + y^2) dxdy

Start by substituting x = rcosΘ and y = rsinΘ

Now find the polar limits of integration for the boundary of R (the area we’re trying to evaluate). Start by sketching the region R and the bounding curve.

Find the R limits by drawing an imaginary ray from the origin through R. Where the ray first enters R is the lower boundary and where the ray leaves R is the upper boundary.

Theta is found as the largest and smallest angle between the imaginary ray and the x-axis that the ray must make to encompass the entirety of region R.

Now replace dxdy with r drdΘ. Note that this extra r will be multiplied by the original equation.

So, with 0 <= Θ <= pi/2 and 0<= r <= 1, we get the polar integral

∫0-pi/2∫0-1 ( (rcosΘ)^2 + (rsinΘ)^2 ) * r drdΘ

Simplifying this using the trigonometric identity sin(x)^2 + cos(x)^2 = 1, we get:

**Integral in Polar Form: **∫0-pi/2 ∫0-1 r^3 drdΘ

Evaluate the integral as follows:

∫0-pi/2 ( r^4)/4 |0-1 dΘ

∫0-pi/2 1/4 dΘ -> Θ/4 |0 – pi/2 = pi/8

**Answer: pi/8**