Multiple Dimensions - Extending Disk Integration

In calculus, disk integration, allows us to calculate the volume of a solid by integrating along an axis. It models the resulting shape as a stack of disks of infinitesimal thickness. Most of the time, we chose an axis of revolution, which results in a circular disk, though this is not necessary. The washer method uses the same idea, and gives us a 'hollow' solid of revolution. We'd push this idea even further.

Let's consider the case of a circle with radius rr. We know that it has an area of πr2 \pi r^2 and a perimeter of 2πr 2 \pi r , but how can we show this?

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One basic approach, is to say that the circle is the set of points such that x2+y2r2 x^2 + y^2 \leq r^2 , and hence we want to calculate R1,dR \int_R 1, dR . Let's use the area element, with a chosen axis of the x x - axis. When x=X x = X , we see that y2r2X2 y^2 \leq r^2 - X^2 and so r2X2yr2X2 - \sqrt{ r^2 - X^2} \leq y \leq \sqrt{ r^2 - X^2} . Hence, at x=X x = X , the area contributed is equal to r2X2(r2X2)=2r2X2 \sqrt{ r^2 - X^2} - (-\sqrt{ r^2 - X^2} ) = 2 \sqrt{ r^2 - X^2} . As such, this allows us to calculate that:

A(x)dx=rr2r2x2dx=πr2 \int A(x) dx = \int_{-r}^r 2 \sqrt{ r^2 - x^2} \, dx = \pi r^2

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Another way to use the area element, is to think about obtaining the area of the circle, by considering how faraway each of the points are. In this case, our chosen axis would be the radius of the circle. When the radius is RR, the contribution to the area is p(R) p(R) , which denotes the perimeter of the circle. This tells us that

0rP(R)dR=πr2. \int_0^r P(R) \, dR = \pi r^2.

Applying the fundamental theorem of calculus, we get that P(r)=2πr P(r) = 2 \pi r . We see this relationship represented in the previous 2 problems, where we worked on the 2 dimensional circle and the 3 dimensional sphere.

This seems like a torturous way to obtain the area and perimeter of a circle, which are already well known. In the following questions, we will apply this method to find the area and perimeter of a sphere in 4 dimensions!

Image credit: Wikipedia Macks, Wikipedia Ksmrq

Note by Calvin Lin
5 years, 5 months ago

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Thanks for a nice presentation. What about the hollow cylinder method ?

Niranjan Khanderia - 5 years, 5 months ago

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Yes, there are also extensions to the shell method. I avoided talking about it, because it's not intuitive and most people have some trouble grasping it. It uses the same idea of "area element", which is how I prefer to think about it.

Calvin Lin Staff - 5 years, 5 months ago

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Thank you. In some case the integral is simple by one method or the other.

Niranjan Khanderia - 5 years, 5 months ago

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0/3

Shakeel Ahmed - 5 years ago

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Nice presentation

Laith Hameed - 5 years, 4 months ago

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i dont know

JayJay Ladiv - 5 years, 4 months ago

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The answer 0/1

Eugene Joseph Rivera - 5 years, 5 months ago

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