| Integrand (x,y) | Region | Converted polar integrand | Known result |
|---|---|---|---|
| x^2 + y^2 | Disk, R=2 | (r^2) * r = r^3 | Exact: 8π ≈ 25.1327412287 |
| 1 | Annulus, 1≤r≤3, 0≤θ≤2π | 1 * r = r | Area: π(3²−1²)=8π ≈ 25.1327 |
Polar conversion uses the substitution: x = r cos(θ), y = r sin(θ). The area element transforms as dx dy = r dr dθ.
If your integrand is f(x,y), the polar integrand becomes f(r cos(θ), r sin(θ)) · r. If your integrand is already g(r,θ), it becomes g(r,θ) · r.
- Enter an integrand in Cartesian or polar variables.
- Select a region type that matches polar bounds.
- Provide r limits and angle limits (use pi if needed).
- Choose radians or degrees for angle bounds.
- Press Convert and Calculate to view results above.
- Use the download buttons for CSV or PDF reports.
Why convert a double integral to polar form
Many regions are naturally circular: disks, rings, and wedges. In Cartesian form the same boundary often becomes multiple curves, but polar bounds can be constant numbers. That reduces setup time and lowers algebra mistakes when integrating over symmetry.
The core substitution used by this calculator
The mapping is x=r cos(θ) and y=r sin(θ). Distances satisfy x²+y²=r², so terms like x²+y² simplify immediately. The area element changes by the Jacobian determinant, giving dx dy = r dr dθ. Because r represents distance from the origin, this form is ideal when boundaries depend on radius.
How integrands change during conversion
If you enter f(x,y), the tool builds f(r cos(θ), r sin(θ)) and multiplies by r. For example, f=x²+y² becomes r², then r²·r = r³. Constants stay constant, and trigonometric factors keep θ explicitly. When your expression already uses r and θ, only the extra r factor is appended.
Choosing bounds for common regions
For a disk of radius R, use 0≤r≤R and 0≤θ≤2π. For an annulus, use rin≤r≤rout with the same θ span. For a sector, keep r limits and restrict θ, such as 0≤θ≤π/3. If your region is described by a curve like r=2cos(θ), set r from 0 up to that curve and choose θ where r stays nonnegative.
Angle units and practical ranges
Angles are usually in radians: 2π≈6.283185. Degrees can be convenient for geometry; 360° corresponds to 2π, 90° to π/2, and 45° to π/4. The calculator converts degree bounds internally before evaluation. Remember that θ is periodic, so intervals like -π to π describe the same full turn as 0 to 2π.
Numerical evaluation and accuracy controls
After conversion, the result is estimated with a grid in r and θ. Increasing Nr and Nθ reduces step sizes, improving accuracy for rapidly changing integrands. Smooth integrands over simple regions converge quickly; sharp peaks need finer grids.
Quick reliability checks with known results
Use test cases to validate inputs. Over a disk R=2, ∫∫(x²+y²)dA equals ∫₀^{2π}∫₀² r³ dr dθ = 8π≈25.132741. Over an annulus 1≤r≤3, ∫∫1 dA equals area π(9−1)=8π. If your numeric output is far from these values, recheck bounds, angle units, and the integrand type.
Do I always multiply by r in polar form?
Yes. The area element changes from dx dy to r dr dθ. So the converted integrand is your substituted expression multiplied by r, even when the original formula already uses r and θ.
Which region option should I pick?
Use Disk when r runs from 0 to a single radius R. Use Annulus when r has inner and outer limits. Use Sector when you also restrict θ to an interval smaller than a full turn.
Can I enter angle bounds in degrees?
Yes. Choose Degrees for θ limits, and the calculator converts them to radians internally. This avoids manual conversions like 180°=π and 360°=2π.
What if my integrand is Cartesian but my bounds are polar?
Select Cartesian integrand type and enter f(x,y). The tool substitutes x=r cos(θ) and y=r sin(θ) automatically, then applies the Jacobian factor r for the final polar integrand.
How do I get more accurate numeric results?
Increase N_r and N_θ so the r–θ grid is finer. Accuracy improves for smooth functions, but steep changes or narrow sectors usually need higher counts and carefully chosen bounds.
Why does my result look wrong or unexpected?
Recheck that r limits are nonnegative and ordered correctly, and confirm your angle unit. Also verify whether symmetry cancels terms over a full 0 to 2π interval, producing a smaller value than expected.