Welcome to four examples
on how to convert a spherical equation
to a rectangular equation.
We are asked to find an equation in rectangular coordinates
for the spherical equation rho equals three,
which is graphed below.
For review, a point in spherical coordinates
is in the form row, theta, phi,
which are pictured here
on the three-dimensional coordinate system.
In rectangular coordinates,
a point is in the form x, y, z,
which means to convert an equation
from spherical coordinates to rectangular coordinates,
we need to perform substitutions for rho, theta, and phi,
so the equation only contains the variables x, y, and z.
To perform these substitutions, we use the equations below
that relate x, y, z, rho, theta, and phi.
So going back to our equation, we have rho equals three,
looking at the equations below,
notice here we have rho squared equals x squared
plus y squared plus z squared.
Instead of solving this equation for rho,
let's square both sides of the given equation,
which gives us the square of rho equals the square of three,
which is rho squared equals nine,
and now we can perform a substitution for rho squared
giving us x squared plus y squared
plus z squared equals nine.
Notice now the equation only contains
the variables x, y, and z,
and therefore we know the equation is in rectangular form.
And we should recognize this equation
as a sphere center of the origin with a radius of three.
Our next spherical equation
is rho equals three cosine phi.
Notice on the left side of the equation,
we just have rho, not rho squared,
let's focus on the right side of the equation.
Let's perform a substitution for cosine phi.
Here, we can see that cosine phi equals z divided by rho.
Or if we have this equation and we solve for cosine phi,
we still have cosine phi equal z divided by rho.
So performing this substitution for a cosined phi,
we have rho equals three
times z divided by rho,
or rho equals three z divided by rho.
For the next step, let's clear this fraction
from the equation by multiplying both sides by rho.
Notice by doing this, we do get rho squared
on the left side of the equation,
which'll make for an easier substitution.
Simplifying, we have rho squared equals,
here rho divided by rho simplifies to one,
the rise head is just three z.
So we have rho squared equals three z,
and again rho squared equals x squared
plus y squared plus z squared,
giving us the rectangular equation
x squared plus y squared
plus z squared equals three z.
We know this is now in rectangular form
because the equation only contains
the variables x, y, and z.
And notice how this is a sphere,
but it is not centered at the origin.
Next, we have the spherical equation
rho equals six cosecant phi
secant theta.
Looking at the equations below,
notice how the trick equations only contain sine and cosine,
though we do have an equation involving tangent here,
let's write cosecant and secant using sine and cosine.
Remember, cosecant is equal to one over sine,
and secant is equal to one over cosine.
So we can write this as
rho equals six times one over sine phi
times one over cosine theta,
or rho equals six over sine phi
cosine theta.
Now let's clear this fraction from the equation
by multiplying both sides of the equation
by the denominator of sine phi cosine theta.
On the left side of the equation,
we now have rho sine phi
cosine theta,
on the right side, this product divided by itself
simplifies to one, we just have six.
And then finally, rho times sine phi
times cosine theta is equal to x.
This simplifies nicely to x equals six,
which we should recognize in three dimensions
is this plane here intersecting the x-axis at six.
Let's look at one more example.
Here we have theta equals pi divided by three.
Here we need to perform a substitution for theta.
We're given tangent theta equals y divided by x,
which means theta is equal to arctangent,
or inverse tangent, of y divided by x.
Performing this substitution gives us
arctangent of y divided by x
equals pi divided by three.
From here, let's undo the arctangent or inverse tangent
by taking the tangent of both sides of the equation.
We would have the tangent
of the inverse tangent or arctangent of y divided by x
equals tangent pi divided by three.
On the left side, the tangent undoes the inverse tangent,
and the left side simplifies nicely to y divided by x.
Tangent pi over three is equal to tangent 60 degrees,
which is equal to the square root of three
divided by one, or square root three.
We can check this on the unit circle,
or using reference triangles.
Remember pi over three
is equal to 60 degrees.
Here's a 30-60-90 right triangle,
where we can label the sides one, two, square root three.
Tangent pi over three or tangent 60 degrees
is equal to the ratio of the opposite side
to the adjacent side,
which is square root three over one, or square root three.
So going back to our equation,
this equation is now in rectangular form
because it only contains the variables x and y.
Let's go ahead and solve this for y.
Solve this for y, we multiply both sides
of the equation by x.
Simplifying we have y equals square root three x,
which, in three dimensions, is the plane graphed below.
I hope you found this helpful.