I was given this problem:
Use double integrals to find the area under the curve defined by $r=1+\sin\theta$.
We can see that $0\leq\theta\leq2\pi,$ and $0\leq r\leq 1+\sin\theta.$ My question is, why do I have to use the Jacobian when originally the curve was given in the coordinates that I want to integrate? To be clearer, if I want to find the area of a circle with radius $a$ which is described by $x^2+y^2=a^2$, then I change variables: $x=r\cos\theta$ and $y=r\sin\theta$. The integral on the circle:
$$\iint\limits_{\text{Circle in $xy$ plane}} 1\,dA = \iint\limits_{\text{Rectangle in $r\theta$ plane}}1·|\boldsymbol J|\,d\hat{A}.$$ However in the problem I'm already given the curve in $r\theta$. Why should I add the Jacobian? I appreciate your thoughts.
The idea of the double integral is to find a small area and integrate it to find the total area. This small area constructed by small changes $dx$ and $dy$ in the coordinates $x$ and $y$, so the small area is given by $dA=dxdy$ while in case of the polar coordinates $r,\theta$ the small changes are $dr$ and $d\theta$ and in this case the small area is $dA=rdrd\theta$
In general we have that $$\int\int f(x,y)dxdy=\int\int g(u,v)J(u,v)dudv$$ where the area element $dA=dxdy$ in Cartesian in the new coordinates is given by $$dA=J(u,v)dudv=\frac{\partial(x,y)}{\partial(u,v)}dudv$$ The Jacobian says how to get the area element using the given transformation. For example in polar coordinates $dA=rd\theta dr$ as in the following figure(I copied it from Google)