# Integrating to find the volume

• Aki
In summary, the conversation is about finding the hypervolume of a hypersphere using the equation V^4= 2(\frac{4\pi} {3}) \int_0^r (\sqrt{r^2-x^2}) ^3 dx. The person is unsure of how to integrate and expand the equation, and asks for help with coordinate conversions. Another person suggests using a trigonometric substitution, but the original person states that the hypersphere has zero volume. The conversation ends with the original person providing the coordinate conversions for the equation.

#### Aki

Help, I'm trying to find the hypervolume of a hypersphere and I'm stuck on this:

$$V^4= 2(\frac{4\pi} {3}) \int_0^r (\sqrt{r^2-x^2}) ^3 dx$$

I don't know how to do the integration, and I can't expand the $$(\sqrt{r^2-x^2}) ^3$$
The answer should be $$\frac{\pi ^2}{2}r^4$$

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can you please state the original problem in it's entirety?

what are your coordinate conversions (i.e. what is $x$ in terms of $r$, $\theta$, etc.)?

often times by converting to other coordinates (hyperspherical coordinates, in this case) the integral becomes simpler.

Anyway,it's useless.The hypersphere has zero volume.

Daniel.

The hypersphere does have a "hyper-volume" (or, more generally, a "measure")!

Aki: in order to integrate $$V^4= 2(\frac{4\pi} {3}) \int_0^r (\sqrt{r^2-x^2}) ^3 dx$$, did you consider using a trigonometric substitution, say x= r sin(&theta;)?

That's the BALL,the hypersphere (the S_{3} being the smallest in dimension) is a hypersurface and has zero hypervolume.

Daniel.

$$x=r \cos \theta , \sqrt {r^2 - x^2} = r \sin \theta , dx = -r \sin \theta d\theta$$

The volume enclosed by a hypersphere.

## What is integration and how is it used to find volume?

Integration is a mathematical technique used to find the area under a curve or the volume of a solid. It involves breaking down a shape into infinitesimally small pieces and adding them up to get the total area or volume.

## What is the formula for finding volume using integration?

The formula for finding volume using integration is ∫a^bA(x)dx, where "a" and "b" represent the limits of integration and A(x) represents the cross-sectional area of the shape at a given x-value.

## What types of shapes can be measured using integration to find volume?

Integration can be used to find the volume of any solid shape, including cones, cylinders, spheres, and irregular shapes. It can also be used to find the volume of 3D objects with changing cross-sectional areas, such as a vase or a bottle.

## How does the precision of integration affect the accuracy of the calculated volume?

The more precise the integration process, the more accurate the calculated volume will be. This means using smaller intervals and increasing the number of intervals to get a more accurate approximation of the shape's volume.

## What are some real-world applications of integration to find volume?

Integration is used in many fields, such as physics, engineering, and architecture, to calculate the volume of objects. It is also used in fluid dynamics to find the volume of liquids and gases in containers or flowing through pipes. Additionally, integration is used in computer graphics to create 3D models and animations.