### Why do you multiply by an extra "r" when converting a double integral in polar coordinates? (Transformation example 4)

Let me answer a commonly asked question, 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, \(x = r\cos (\theta )\) and \(y = r\sin (\theta )\), 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!

Please see our other examples on finding the Jacobian for a multivariate transformation:

Jacobian of a 3×3 Matrix Transformation (Ex5)

Alternatively return to the Jacobian hub:

Jacobians for Transformations