In my textbook the area if a washer is appropriated by a washer is $2\pi r_i\Delta r$ while what I know the area if a washer to be is $\pi(r_i + r_{i-1})\Delta r.$
Here's the segment of my textbook that I'm confused about:
In Example 7 in Section 3.7 we discussed the law of laminar flow:$$v(r)=\dfrac P{4\eta l}(R^2-r^2)$$which gives the velocity $v$ of blood that flows along a blood vessel with radius $R$ and length $l$ at a distance $r$ from the central axis, where $P$ is the pressure difference between the ends of the vessel and $\eta$ is the viscosity of the blood. Now, in order to compute the rate of blood flow, or flux (volume per unit time), we consider smaller, equally spaced radii $r_1,r_2,...$ The approximate area of the ring (or washer) with inner radius $r_{i-1}$ and outer radius $r_i$ is$$2\pi r_i\Delta r\quad\text{where}\quad\Delta r=r_i-r_{i-1}$$(See Figure 4.) If $\Delta r$ is small, then the velocity is almost constant throughout this ring and can be approximated by $v(r)$. Thus the volume of blood per unit time that flows across the ring is approximately$$(2\pi r_i\Delta r)v(r_i)=2\pi r_iv(r_i)\Delta r$$
The exact value of the area is $$ \pi r_i^2 - \pi r_{i-1}^2 = \pi (r_i^2 - r_{i-1}^2) = \pi (r_i + r_{i+1}) (r_i - r_{i-1}) = 2\pi \left(\frac{r_i + r_{i+1}}{2}\right)\Delta r $$ As $\Delta r \to 0$, the difference between the average radius $\frac{r_i + r_{i+1}}{2}$ and $r_i$ tends to zero as well. So we can make this substitution as an approximation.