### Home > CALC > Chapter 2 > Lesson 2.2.1 > Problem2-55

2-55.

Base of each rectangle:

$\Delta x=\frac{\pi }{6}$

Height of each rectangle is determined by the function: f(x) = sinx evaluated at each midpoint.

Let's examine one of the rectangles, the smallest one:

$\text{Midpoint of smallest rectangle: }x= \frac{1}{2}\left ( \frac{\pi }{6} \right )=\frac{\pi }{12}$

$\text{Height of smallest rectangle: }y= \text{sin}\frac{\pi }{12}$

$\text{Area of smallest rectangle: }A= bh=\left ( \frac{\pi }{6} \right )\left ( \frac{\pi }{12} \right )=\left (\frac{\pi }{72} \right )$

Two medium sized rectangles with midpoints at ____ and ____.
Two large rectangles with midpoints at ____ and ____.
Find the area of all six rectangles.

Notice there are two small rectangles with midpoints at

$\frac{\pi }{12}\text{ and } \frac{11\pi }{12}.$

Imagine a barbell with six cylindrical weight: two small, two medium and two large. Each cylinder has a thickness (or height) of

$\frac{\pi }{6},$

but their circular bases vary in sizes. The radius of each circular base is determined by the function f(x) = sin x evaluated at

$x=\frac{\pi }{12}, \frac{3\pi }{12}, \frac{5\pi }{12}, \frac{7\pi }{12}, \frac{9\pi }{12}, \frac{11\pi }{12}.$

The smallest cylinders are on the outside, then medium then large.

Because of symmetry, you can evaluate the volume of three of the cylinders and then double the result.

Volume of a cylinder: V= πr²h = πr²(Δx)