Home»Math Guides»Double Iterated Integrals using Polar Coordinates (circular region of integration)
Why do you multiply by an extra “r” when converting a double integral in polar coordinates?
Let me answer a commonly asked question, “Why do you need to always multiply by an extra “r” term when converting to polar coordinates?”
The answer is simple, it’s because it’s a transformation of multiple variables (2 variables to 2 new variables).
Remember when you did substitution involving 1 variable to 1 new variable? For example, if you let \(u = {x^2}\), then \(du = 2xdt\).
But if you’re using polar coordinates, which is, \(x = rcos (\theta )\) and \(y = rsin (\theta )\), then you need to find the Jacobian, which is the determinant of all the partial derivatives.
You can’t just differentiate it as before to change the variable.
\[J = \left| {\begin{array}{*{20}{c}}{\frac{{\partial x}}{{\partial r}}}&{\frac{{\partial x}}{{\partial \theta }}}\\{\frac{{\partial y}}{{\partial r}}}&{\frac{{\partial y}}{{\partial \theta }}}\end{array}} \right|\]
\[J = \left| {\begin{array}{*{20}{c}}{\cos (\theta )}&{ – r\sin (\theta )}\\{\sin (\theta )}&{r\cos (\theta )}\end{array}} \right|\]
\[J = \cos (\theta ) \cdot r\cos (\theta ) – ( – r\sin (\theta )) \cdot \sin (\theta ))\]
\[J = r{\cos ^2}(\theta ) + r{\sin ^2}(\theta )\]
And recall the identity:
\[{\cos ^2}(\theta ) + {\sin ^2}(\theta ) = 1\]
Simplifying our Jacobian:
\[J = r\]
What exactly does the Jacobian mean in the transformation?
It’s just the factor you need to multiply by when you do a transformation change of coordinates.
In other words,
\[dxdy = rdrd\theta \]
And that’s why you multiply by an extra “r” term when you do a polar coordinates transformation!
Try the next example problem where you solve a double integral using polar coordinates!