Multiple Dimensions - Volume Of a 4-Sphere

Calculus Level 3

Let's consider how to find the volume of the sphere with radius r r in 4 dimensions. The sphere is the set of points such that x 1 2 + x 2 2 + x 3 2 + x 4 2 r 2 . x_1 ^2 + x_2 ^2 + x_3 ^2 + x_ 4 ^2 \leq r^2.

Let's calculate the volume by integrating along the x 1 x_1 axis. The volume is equal to V ( r ) = r r A ( x 1 ) d x 1 V(r) = \int_{-r} ^{r} A(x_1) \, d x_1 , where A ( x 1 ) A(x_1) denotes the area element for a given value of x 1 x_1 .

What is the area element A ( x ) A(x) and what is the corresponding volume V ( r ) V(r) of the sphere ?

Image credit: Wikipedia Pbroks13
A ( x ) = 4 3 π r 3 , V ( r ) = 1 2 π 2 r 4 A(x) = \frac{4}{3} \pi r^3 , V(r) = \frac{ 1}{2} \pi ^2 r^4 A ( x ) = 4 3 π ( r 2 x 2 ) 3 , V ( r ) = 1 2 π 2 r 4 A(x) = \frac{4}{3} \pi ( \sqrt{ r^2 - x^2 } ) ^3 , V(r) = \frac{ 1}{2} \pi ^2 r^4 A ( x ) = 4 π r 3 , V ( r ) = 3 2 π 2 r 4 A(x) = 4 \pi r^3 , V(r) = \frac{ 3}{2} \pi ^2 r^4 A ( x ) = 4 3 r π ( r 2 x 2 ) , V ( r ) = 16 9 π r 4 A(x) = \frac{4}{3} r \pi ( r^2 - x^2 ) , V(r) = \frac{ 16}{9} \pi r^4

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2 solutions

Chung Kevin
May 15, 2014

When x 1 = x x_1 = x , then we have x 2 2 + x 3 2 + x 4 2 r 2 x 2 x_2 ^2 + x_3^2 + x_4 ^2 \leq r^2 - x^2 , which is a sphere of radius r 2 x 2 \sqrt{ r^2 - x^2 } . Hence, the corresponding area element would be equal to the volume of the sphere, thus A ( x ) = 4 3 π R 3 = 4 3 π r 2 x 2 3 A(x) = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \sqrt{ r^2 - x^2 } ^3 .

To find the volume of the sphere, we need to integrate this. We have

V ( x ) = r r 4 3 π r 2 x 2 3 d x V(x) = \int_{-r} ^ r \frac{4}{3} \pi \sqrt{ r^2 - x^2 } ^3 \, dx

We can use the substitution x = r cos θ x = r \cos \theta , d x = r sin θ d θ dx = r - \sin \theta \, d \theta , we get

4 3 π π 0 r 3 sin 3 θ × ( r sin θ ) d θ = 4 3 π × 3 π r 4 8 = 1 2 π 2 r 4 \frac{4}{3} \pi \int_{\pi} ^ 0 r^3 \sin^3 \theta \times ( - r \sin \theta) \, d \theta = \frac{4}{3} \pi \times \frac{ 3 \pi r^4 }{8} = \frac{ 1}{2} \pi^2 r^4

( I skipped sin 4 θ d θ \int \sin^4 \theta\, d \theta which you can do by parts.)

Thanks and congrats.

Niranjan Khanderia - 7 years ago
Kartikey Kumar
May 15, 2014

(Fun in multiple dimensions == f.i.m.d)

From previous questions of f.i.m.d we concluded derivative of volume of sphere equals surface area and derivative of area of circle equals perimeter.

by similar arguments, derivative of volume in 4d equals surface area. It means that the derivative of answer of item 6 f.i.m.d question must be equal to volume. Differentiating each option of item 6 f.i.m.d question and comparing with volume options in this question we get

surface area =2((pi)^2)r^3 and volume= 1/2 ((pi^2))r^4

Since integration of A(x)dx must equal to V(x) we conclude that A(x)=(4/3)(pi)(r^2-x^2)^(3/2)

Note that you are not doing the problems in sequence. We use this item, to justify the answer of item 6.

Also, note that there are 2 options with that give the same volume. Hence, you need to justify why the area element A ( x ) A(x) is what you claim it to be.

Calvin Lin Staff - 7 years ago

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It can be justified that the other option is wrong as "Integral of A(x)dx(from 0 to r) " must equal to"V(x)(at r)" this is not true for the wrong option. Although my answer is just simple reasoning. But mathematically Chung kevin solved the problem very well.

Kartikey Kumar - 7 years ago

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