Integral To Polar Coordinates Calculator

The Integral to Polar Coordinates Calculator helps convert and evaluate integrals from Cartesian coordinates to polar coordinates. This is useful for simplifying the evaluation of certain integrals, especially in cases with circular symmetry.

Inputs:

1. Cartesian Function (f(x, y)): The function to be integrated in Cartesian coordinates.
2. Limits for x (xmin,xmaxx_{min}, x_{max}xmin​,xmax​): The range of the x-variable.
3. Limits for y (ymin,ymaxy_{min}, y_{max}ymin​,ymax​): The range of the y-variable.

Formulas and Calculations:

Conversion to Polar Coordinates:

The conversion formulas from Cartesian coordinates (x,y)(x, y)(x,y) to polar coordinates (r,θ)(r, \theta)(r,θ) are: x=rcos⁡(θ)x = r \cos(\theta)x=rcos(θ) y=rsin⁡(θ)y = r \sin(\theta)y=rsin(θ) dx dy=r dr dθdx \, dy = r \, dr \, d\thetadxdy=rdrdθ

Integral Conversion:

The double integral in Cartesian coordinates: ∬Rf(x,y) dx dy\iint_R f(x, y) \, dx \, dy∬R​f(x,y)dxdy is converted to polar coordinates as: ∬Rf(rcos⁡(θ),rsin⁡(θ)) r dr dθ\iint_R f(r \cos(\theta), r \sin(\theta)) \, r \, dr \, d\theta∬R​f(rcos(θ),rsin(θ))rdrdθ

Example Calculation:

Inputs:

• Cartesian Function: f(x,y)=x2+y2f(x, y) = x^2 + y^2f(x,y)=x2+y2
• Limits for xxx: −1≤x≤1-1 \leq x \leq 1−1≤x≤1
• Limits for yyy: 0≤y≤1−x20 \leq y \leq \sqrt{1 – x^2}0≤y≤1−x2​

Step-by-Step Calculation:

1. Express the Function in Polar Coordinates: f(x,y)=x2+y2f(x, y) = x^2 + y^2f(x,y)=x2+y2 f(r,θ)=(rcos⁡(θ))2+(rsin⁡(θ))2=r2cos⁡2(θ)+r2sin⁡2(θ)=r2(cos⁡2(θ)+sin⁡2(θ))=r2f(r, \theta) = (r \cos(\theta))^2 + (r \sin(\theta))^2 = r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = r^2 (\cos^2(\theta) + \sin^2(\theta)) = r^2f(r,θ)=(rcos(θ))2+(rsin(θ))2=r2cos2(θ)+r2sin2(θ)=r2(cos2(θ)+sin2(θ))=r2
2. Set up the Integral in Polar Coordinates:
   • The given region in Cartesian coordinates is a semicircle of radius 1 centered at the origin.
   • The limits in polar coordinates are: 0≤r≤10 \leq r \leq 10≤r≤1 0≤θ≤π0 \leq \theta \leq \pi0≤θ≤π
3. Convert the Integral: ∬Rf(x,y) dx dy=∫0π∫01r2⋅r dr dθ\iint_R f(x, y) \, dx \, dy = \int_0^\pi \int_0^1 r^2 \cdot r \, dr \, d\theta∬R​f(x,y)dxdy=∫0π​∫01​r2⋅rdrdθ
4. Evaluate the Inner Integral (with respect to rrr): ∫01r3 dr=[r44]01=14\int_0^1 r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^1 = \frac{1}{4}∫01​r3dr=[4r4​]01​=41​
5. Evaluate the Outer Integral (with respect to θ\thetaθ): ∫0π14 dθ=14[θ]0π=π4\int_0^\pi \frac{1}{4} \, d\theta = \frac{1}{4} \left[ \theta \right]_0^\pi = \frac{\pi}{4}∫0π​41​dθ=41​[θ]0π​=4π​

Conclusion:

For the function f(x,y)=x2+y2f(x, y) = x^2 + y^2f(x,y)=x2+y2 over the given region, the integral in Cartesian coordinates converts to an integral in polar coordinates and evaluates to π4\frac{\pi}{4}4π​.

These calculations help engineers, mathematicians, and students convert and evaluate integrals more easily when dealing with circular or radial symmetry.