Breaking The Disc Method

Calculus Level 3

The above figure consists of a rectangle with a semicircle cut out of one end and added to the other end, where $L$ is the width of the rectangle, and the curved length of a semicircle is $\pi r$ .

To calculate the area of the shaded figure, Svatejas applies the disc method as follows:

Consider the axis of integration to be the semicircular arc, which has length $\pi r$ . For each horizontal strip, we have an area element (technically length element) of $L$ . Hence, the area is

$\int_{R} L \, dR = \pi r \times L.$

What is the area of the shaded figure?

Inspiration, see solution comments .

2 \pi r L

\pi r L

r L

2 r L

1 solution

Calvin Lin Staff
Jan 19, 2016

[This is not a solution. It is an explanation of why the fallacy resulted.]

For those who learnt about disc method to find the volume / area of the element, what is wrong with the following argument :

Consider the axis of integration to be the semicircular arc, which has length $\pi r$ . For each horizontal strip, we have an area element (technically length element) of $L$ . Hence, the area is

$\int_{R} L \, dR = \pi r \times L$

Moderator note:

Try spotting the fallacy before reading the comments.

The issue with the argument is that we are summing up the area elements incorrectly . Think back to the integration that you learnt. Why is integration in cartesian coordinates so different from integration in polar coordinates? In particular, why is $1 \, dy \, dx = r \, dr \, d\theta$ ? Wouldn't it be simplier if the area was just $1\, dr \, d\theta$ ? (Yes, I know many students wish for that) This suggests that the angle between these axis has an effect on the calculated volume / area.

Consider the following problem: If we introduce new tilted coordinates $a , b$ , where the angle between $(0,1)$ and $(1, 0)$ is $60 ^ \circ$ , what is the area of the quadrilateral bounded by $(0,0), (0,1), (1,1), (1,0)$ ?

Why isn't the area of this "rectangle" just 1? Shouldn't it be $\int_0^1 1 \, dx$ ? The reason why we can't just integrate $\, dx$ , is because the area element that we chose is not perpendicular to our axis of integration, and we need to account for the tilt. In this case, it's easy. We want it to be $\int_0^1 \, ( \sin \theta \, dx )$ , and so we arrive at $\sin 60 ^ \circ$ .

Similarly in this case, the area element that you chose is in not in the perpendicular direction compared to your axis of integration, and hence the area isn't just $\int_0^{\pi r } L \, dR$ .

Extension: What do you think is the general formula? Hint: dot product .

Calvin Lin Staff - 5 years, 4 months ago

Breaking The Disc Method

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