关于将二重积分转换为极坐标时被积函数形式的疑问（Ron Larson《微积分》第8版第14章第3节第15题）

Hey there! Let me break down where that extra r comes from—it's a super common point of confusion when first learning polar coordinates for double integrals, so you're definitely not alone here.

When converting a double integral from rectangular coordinates to polar coordinates, it's not just about swapping $x = r\cos\theta$ and $y = r\sin\theta$ in the integrand. We also have to account for how the area element changes between the two coordinate systems:

• In rectangular coordinates, the tiny area element is $dA = dx , dy$.
• In polar coordinates, that same tiny area isn't a square anymore—it's a small wedge shape. The correct area element here is $dA = r , dr , d\theta$. That extra r is called the Jacobian determinant, which acts as a scaling factor for area when you switch coordinate systems.

So applying this to your problem:

1. Your original integrand is $y$, which translates to $r\sin\theta$ in polar coordinates.
2. Then you multiply by the Jacobian factor r from the area element.
3. That gives you $r\sin\theta \times r = r^2\sin\theta$, which matches the solution you saw.

A quick intuitive way to remember this: imagine a small ring at radius $r$ with thickness $dr$ and angle $d\theta$. The length of the arc is $r d\theta$, so the area of the tiny wedge is roughly $r d\theta \times dr = r , dr , d\theta$—that's where the r comes from!

Don't worry, this is a super easy step to miss at first. Once you do a few more problems, remembering to include that Jacobian factor will become second nature.

备注：内容来源于stack exchange，提问作者naisuu42