Homework Question -Vector Calculus Area I want to calculate the area of a semi-circle. I can use this $\iint x^2+y^2\,\mathrm dx\,\mathrm dy$ or I can use this $\iint r^2 r^2 \sin(\phi) \,\mathrm \,d\theta\,\mathrm dr$. I can see where the $r\,\mathrm d\theta \, dr$ comes from but why is it wrong to say $x=r\cos\theta$ and $y=r\sin \theta$, then differentiate and multiply and get $$(dr)^2\cos\theta \sin\theta +r\,dr\,d\theta \cos^2 \theta −r\,dr\,d \theta \sin^2\theta −r^2(d\theta)^2\sin\theta \cos\theta?$$
Obviously I can't use the above in the integral but I cant see where this is wrong.
 A: A product of differentials is anti-symmetric. That is, $dr\,d\theta= -d\theta \,dr$.  It is also follows that $dr\,dr= -dr\,dr$ so that $dr\,dr= 0$ and $d\theta \,d\theta= -d\theta \,d\theta$ so that $d\theta \,d\theta= 0$.
If $x= r\cos(\theta)$ and $y= r\sin(\theta)$
then $dx= \cos(\theta)\,dr- r \sin(\theta)\,d\theta$ and
$dy= \sin(\theta)\,dr+ r \cos(\theta)\,d\theta$ so that
\begin{align}
dx\,dy = {} & (\cos(\theta)\,dr- r\sin(\theta)\,d\theta)(\sin(\theta)\,dr+ r \cos(\theta)\,d\theta) \\[8pt]
= {} & \cos(\theta)\sin(\theta)\,dr\,dr+ r\cos^2(\theta) \, dr \, d\theta \\
& {} - r \sin^2(\theta)\,d\theta \,dr- r^2 \sin(\theta) \cos(\theta) \,d\theta \,d\theta \\[8pt]
= {} & 0+  r \cos^2(\theta)\,dr \,d\theta+ r \sin^2 \,dr\,d\theta+ 0 \\[8pt]
= {} & r \,dr \,d\theta
\end{align}
A: Firstly, you have the wrong formula. The area of the semicircle in Cartesian coordinates would be
$$\int_{-r}^r \sqrt{r^2-x^2} \ dx$$
or, as a double integral,
$$\int_{-r}^r \int_0^{\sqrt{r^2-x^2}} \,dy \ dx$$
Secondly, you're not following the process for changing variables in double integrals, given on the linked source as
$$\iint\limits_R f(x,y) \, dx \, dy = \iint\limits_S f[x(u,v),y(u,v)] \left| \frac{\partial(x,y)}{\partial(u,v)} \right| \, du\,dv$$
which ultimately leads to your confusion.
When you change variables in double integrals, you need to compute the Jacobian of your transformation. That means that when you substitute $x=r\cos \theta$ and $y=r\sin \theta$, you need to compute
$$\left| \frac{ 
\partial(x,y)}{
\partial(r, \theta)}
 \right| =\left| \begin{matrix}
\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\
\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}
\end{matrix} \right|$$
In this case, that gives you
$$\left| \begin{matrix}
\cos \theta & -r\sin\theta \\
\sin \theta & r\cos\theta
\end{matrix} \right| =r(\cos^2\theta+\sin^2\theta)=r$$
Once you have this, you can safely convert the integral into polar coordinates to get
$$\int_{-r}^r \int_0^{\sqrt{r^2-x^2}}dy \ dx=\int_0^\pi \int_{0}^r r \ dr \ d\theta=\frac{\pi r^2}{2}$$
