Hello, i am having a problem at understanding the substitution method for a triple integration

Say, i have a triple integral (x^2+y^2+z^2)dxdydz, on a volume V: x^2+y^2+z^2<=a^2

I know the volume refers to the inside of a sphere, with a radius of a, however what i do not understand is why, in order to solve the integration, the following substitutions are made

x= r*cos(t)*cos(f)

y= r*sin(t)*sin(f)

z= r*sin(f)

with r belonging to [0,a]

t belonging to [0,2pi]

f belonging to [-pi/2, pi/2]

I cannot grasp the concept why x,y,z would be equal to sine and cosine function.

I know the substitution is made because it will result in sin^2(x) + cos^2(x) = 1, making the integration easy to solve, but i cannot understand on what basis it is made, so an explanation on that would be much appreciated, thank you

Why is it necessary to also insert the cos(f) and sin(f) functions into there, doesnt cos, sin(t) cover the whole surface already? Please explain.

What you're doing there is changing to spherical co-ordinates. You have marked the co-ordinate system a little differently then what convention states.

In terms of calculating volume, no matter what co-ordinate set you use, think about what you your integrals represent and see if you are actually calculating the volume.

For example,

In the case of a sphere we have phi, theta and p. What are these things? Well P is the distance of travel. So if we are going from the origin to the edge of a sphere, our distance is from 0 to the radius. If you draw this out, this will actually create a line that you can draw an angle to from the z axis, this angle is denoted by phi. So if we take this line and go from 0 to pi, we are covering the volume inside the sphere along this one line onle. But we need the volume of the entire sphere, which can be easily done if we rotate 360 degrees!

Knowing this, we can find out where the following come from

\(\displaystyle x= r*cos(t)*cos(f) \)

\(\displaystyle y= r*sin(t)*sin(f) \)

\(\displaystyle z= r*sin(f) \)

We will start with Z and work our way back.

Draw the triangle from the line 0 to radius a to the z axis and denote phi as the angle between the z axis and the line 0 to a. This will produce the relationship (by right angle triangle rules)

\(\displaystyle \phi = cos^{-1} { \frac{z}{a} } \)

So,

\(\displaystyle z = acos \phi \)

Now draw the triangle from the line 0 to radius a to the xy plain and denote theta as the angle between the x-axis and the projection of the line 0 to a on the xy plane.

This will produce,

\(\displaystyle \theta = tan^{-1} \frac{y}{x} \)

\(\displaystyle x tan \theta = y \)

\(\displaystyle x sin \theta = y cos \theta \)

We also know,

\(\displaystyle r = \sqrt {x^2 + y^2 + z^2 } \)

Solve the 2 equations for x and y (after subbing in z) and you will find the above definitions to be true

Look at my notes about sphereical co-ordinates here:

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So for spherical we would set up the volume of a sphere like

\(\displaystyle \iiint_{Sphere} dV = \int_0^{ 2 \pi } d \theta \int_0^{ \pi } sin \phi d \phi \int_0^a p^2 dp \)

It's important to note that we can also compute this in cylindrical co-ordinates. In the example given in my notes what we end up with is a very very hard integration. The point was to demonstrate how to use spherical co-ordinates, but that problem would actually be best computed with cylindrical co-ordinates!