Certainly not calculus

Calculus Level 3

At points on a circle of radius a, line segments are drawn perpendicular to the plane of the circle, the perpendicular at each point P being of length ks, where s is the length of the arc of the circle measured counterclockwise from (a, 0) to P and k= 1 π 2 a 2 \frac{1}{\pi^2a^2} , as shown here. Find the area of the surface formed by the perpendiculars along the arc beginning at (a, 0) and extending once around the circle.


The answer is 2.

Benny Joseph by Benny Joseph , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Leonel Castillo
Jun 3, 2018

Imagine that you glue the initial point of the surface to the floor and then grab the upper corner and pull on it until the figure is completely stretched and contained in a 2D plane. In this new plane the x − x- axis is the arclength of the circle, and the height of the object at every point is a function of this, namely x π 2 a 2 \frac{x}{\pi^2 a^2} . Thus to get the area under it in this coordinate system just compute ∫ 0 2 π a x π 2 a 2 d x = 4 π 2 a 2 2 π 2 a 2 = 4 2 = 2 \int_0^{2\pi a} \frac{x}{\pi^2 a^2} dx = \frac{4 \pi^2 a^2}{2 \pi^2 a^2} = \frac{4}{2} = 2 .

The bounds of integration are 0 0 and 2 π a 2\pi a because you start measuring at 0 0 and the circumference of this circle (it's total arclength) is 2 π a 2 \pi a .

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Benny Joseph
May 27, 2018

Consider the two images. They both perfectly fit to form a surface of a cylinder. Required Area = ( 2 π a ) ( 2 π a k ) 2 \frac{ (2\pi a)(2\pi a k)}{2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...