Integrating area of a disk using rectangles and triangles I know there are other ways to solve this, but I think the following method should work for calculating the area of a disk: at every integration step I would sum the area of a rectangle and half of a triangle as shown in the attached picture. This led me to this formula: 
$$ 4 * \int_0^r (x*dy + \frac{dx*dy}{2}) = $$ 
$$ = 4 * \int_0^r(x*\frac{dy}{dx} + \frac{dy}{2}) * dx = $$
$$ = 4 * \int_0^r(x*\frac{dy}{dx})*dx + \int_0^r(\frac{1}{2}*\frac{dy}{dx})*dx*dx $$
I am unsure how to proceed solving this though. Am I on the right track? I understand that $\frac{dy}{dx}$ is the derivative of the circle but how do I go about solving $\int_0^r(\frac{1}{2}*\frac{dy}{dx})*dx*dx$ with regards to the $dx*dx$ part?

 A: I think your concept is correct. When we do the summation, the formula and the drawing is good. But it should be $\Delta$ not $d$. It, in fact, is broadly used in numerical methods. 
When it goes to $dx$,$dy$, only the first order term is kept, and $dx \cdot dy$ disappear.
Please refer to the definition of $dx$. Hope it helps.
A: If you label the small distances in the figure $\Delta x$ and $\Delta y$
instead of $dx$ and $dy$, then the figure
actually illustrates a reasonably good idea for a way
to numerically compute the area of a region.
Instead of a simple Riemann sum (which would involve only rectangles)
you have something resembling an application of the trapezoid method 
(and if you were using these same shapes to integrate with respect to $y$ 
then I think it would be an application of the trapezoid method).
Here's what the summation looks like if we set $n + 1$ points 
$(x_0,y_0),(x_1,y_1),\ldots,(x_i,y_i),\ldots,(x_n,y_n)$ along the quarter-arc
of the circle from the positive $y$-axis to the positive $x$-axis,
so that $(x_0,y_0) = (0,r)$ and $(x_n,y_n) = (r,0)$,
and define $\Delta x_i = x_i - x_{i-1}$ and $\Delta y_i = y_i - y_{i-1}$.
We can write the area $A$ as
$$\begin{eqnarray}
A &=&
 4  \sum_{i=1}^n \left(x_{i-1}\;(-\Delta y_i) 
                        + \frac{(\Delta x_i)(-\Delta y_i)}{2}\right) \\
& =& -4 \sum_{i=1}^n\left(x_{i-1} \frac{\Delta y_i}{\Delta x_i} 
        - \frac{\Delta y_i}{2}\right)  \Delta x_i \\
& =& -4 \sum_{i=1}^n\left(x_{i-1} \frac{\Delta y_i}{\Delta x_i}\right)\Delta x_i
 -4 \sum_{i=1}^n\left(\frac12\frac{\Delta y_i}{\Delta x_i}\right)(\Delta x_i)(\Delta x_i)
\end{eqnarray}$$
Notice that I write $(-\Delta y_i)$ rather than $(\Delta y_i)$ where you wrote $dy$.
That's because as $x$ increases, $y$ decreases, so you should define things so that
$\frac{\Delta y_i}{\Delta x_i} < 0$.
Letting $\Delta x_i > 0$ and  $\Delta y_i < 0$ works.
But of course the base$\times$height formula has to use the magnitudes of those
lengths, that is, we want to multiply two positive numbers.
Now consider what happens if you increase $n$ indefinitely, adding more points in such
a way that the maximum of $\Delta x_i$ converged to zero.
This is essentially turning the sum into a definite integral,
as you originally wanted to do.
Writing the maximum of $\Delta x_i$ as
$$\Delta x_{max} = \max_{1\leq i\leq n}\Delta x_i,$$ we have
$$
0 < -\sum_{i=1}^n\left(\frac12\frac{\Delta y_i}{\Delta x_i}\right)(\Delta x_i)(\Delta x_i)
  < - \sum_{i=1}^n\left(\frac12\frac{\Delta y_i}{\Delta x_i}\right)
        (\Delta x_i)(\Delta x_{max})
  = \frac12 r\; (\Delta x_{max}).
$$
That is, as $\Delta x_{max}$ goes to zero, so does the sum on the left.
This tells us that the area will simply be the limit of the other sum, that is,
$$
A = \lim_{\Delta x_{max} \to 0}
        - \sum_{i=1}^n\left(x_{i-1} \frac{\Delta y_i}{\Delta x_i}\right)\Delta x_i
   = -\int_0^r\left(x \frac{dy}{dx}\right)dx.
$$
Since $y$ goes from $r$ to $0$ as $x$ goes from $0$ to $r$,
if we do a change of variables so that we integrate with respect to $y$
instead of $x$, the integral comes out to $\int_0^r x\;dy.$
A: If you did it this way, you would be multiply counting most of the area of every rectangle.
Picture moving $x$ to the right a little; you are proposing adding that new rectangle to the integral, but the bulk of that new rectangle overlaps with the rectangle in the original picture.
Also, each triangle is not exactly a triangle, it's hypotenuse is actually a small arc.
You can tell your approach won't work well from the fact that in the integral you have one term with only one infinitessimal $dx$, and another term with $dxdy$; these can't mix additively to give a meaningful expression.
Sorry, this approach we not pay off for you.
