Welcome to a lesson on converting between
spherical and rectangular equations.
For review, in a spherical coordinate system,
a point P in space is represented
by the ordered triple, rho comma theta comma phi,
where rho is the distance between the point and the origin,
this green distance here,
theta is the angle counter clockwise from the pole,
or positive x-axis in the xy-plane,
which is this blue angle here,
and phi is the angle between the positive z-axis
and the point which is this red angle here.
When converting between the two coordinate systems
we use the equations shown here.
Let's take a look at our first example.
We're asked to convert the rectangular equation x equals two
to spherical form.
The graph of x equals two is shown here below,
which is this plane in 3D.
To write the equation in spherical form,
we need to eliminate the variable x,
and only use the variables rho, theta, and phi.
And we use this first equation
to perform a substitution for x
where x equals rho sine phi cosine theta.
Performing the substitution for x
gives us rho sine phi cosine theta equals two,
and now we do have a spherical equation
because the equation only contains rho, phi, and theta.
Let's go ahead and solve the equation for rho
by dividing both sides by sine phi cosine theta.
Simplifying, on the left, the quotient simplifies to one,
giving us rho equals two divided by the product
of sine phi and cosine theta.
So we can go ahead and leave the equation in this form here,
but we could also use our trig identities,
since one over sine phi is equal to cosecant phi,
and one over a cosine theta is equal to secant theta.
We could also write the spherical equation as
rho equals two cosecant phi secant theta.
Either of these last two equations would be correct.
Next, we're asked to write the equation
x squared plus y squared minus c squared equals zero
in spherical form.
The graph is the double cone shown here below.
We could perform substitutions for x, y, and z,
using these three equations here,
which I do show in an example video,
however, also notice that rho squared is equal to
x squared plus y squared plus z squared.
So let's work on getting
x squared plus y squared plus z squared
on the left side of this equation.
Let's first undo the negative z squared
by adding z squared to both sides.
And then we'll add another z squared to both sides,
so the left side becomes
x squared plus y squared plus z squared.
So we'll have x squared plus y squared,
and then we'll add an extra z squared.
And then on the right side,
we're first going to add z squared
to undo the negative z squared on the left,
which gives us z squared,
and then we'll add another z squared of the right side
because we added this z squared here.
So again we first moved the negative z squared
to the right side which gives us this z squared,
and then we added another z squared to both sides in red.
And now the left side,
x squared plus y squared plus z squared
is equal to rho squared
which gives us rho squared is equal to two z squared,
and since z is equal to rho cosine phi,
z squared is equal to rho squared cosine squared phi,
which gives us rho squared
equals two rho squared cosine squared phi.
And now let's divide both sides by rho squared.
Simplifying, the left side simplifies to one,
we have one equals two.
This quotient simplifies to one,
the right side is two cosine squared phi.
And now let's go ahead and solve for cosine squared phi,
by dividing both sides by two.
Simplifying, let's change the order of the equation,
and write this as cosine squared phi equals 1/2.
So it's probably best to stop here,
and give the equation in spherical form
as cosine squared phi equals 1/2.
However we could also try to solve the equation for phi.
To do this, the next step would be
to take the square root of both sides of the equation,
so let's also show this.
If we square root both sides of the equation,
we have the square root of cosine squared phi
equals plus or minus the square root of 1/2
which simplifies to plus or minus one
divided by the square root of two.
So simplifying we have the square root
of cosine squared phi is cosine phi.
This is equal to plus or minus one
divided by a square root of two,
or if we rationalize the denominator,
we would have plus or minus square root two over two.
The problem with this though is we'll have to find
two angles for phi that satisfy the equation,
one that has a cosine function value
of positive square root of two divided by two,
and another angle that has a cosine function value
of negative square root of two divided by two.
So if we take a look at the unit circle,
notice cosine pi over four is equal to
the square root of two divided by two.
Remember, on the unit circle,
x is equal to the cosine function value,
and since cosine three pi over four is equal to
negative square root two divided by two,
in order to have an equivalent equation
to the original equation, solve for phi,
we have to have two equations,
phi equals pi over four radians,
and phi equals three pi divided by four radians.
So it's probably best to leave the equation
as we have it in this form here.
If we take a look at this graphically,
notice the graph of phi equals pi over four radians
gives us the top of the cone,
and it takes the second equation
of phi equals three pi divided by four radians
to get the full cone, or the top and the bottom of the cone.
And now let's take a look at two examples
of converting from spherical form to rectangular form.
Let's first convert the equation
theta equals pi over four to rectangular form.
The graph of this is the plane shown here below.
Notice tangent theta's equal to y divided by x,
so starting with that equation,
since we know theta is equal to pi over four,
we can substitute pi over four for theta,
which gives us tangent pi over four
is equal to y divided by x,
and tangent of pi over four is equal to one,
giving us one equals y divided by x.
To solve the equation for y, or to clear the fraction,
we multiply both sides by x,
which gives us x equals, simplifying here on the right,
we have y, x equals y, or y equals x,
would be the rectangular equation
for the given spherical equation.
Let's look at one more example.
We're asked to convert the spherical equation
rho equals two cosine phi to rectangular form.
The graph is the sphere shown here below.
To write the equation in rectangular form,
we need to eliminate rho and phi,
and only have the variables x, y, and z.
Looking at this last equation here,
if we solve for a cosine phi,
we would divide both sides by rho,
giving us cosine phi is equal to z divided by rho.
Let's replace cosine phi with z divided by rho.
This gives us rho equals two times z divided by rho,
and now to clear rho from the denominator,
we would multiply both sides of the equation by rho,
which gives us rho squared equals rho divided by rho,
simplifies to one here, the right side simplifies to two z,
and now we can eliminate the rho squared
by using this equation here,
rho squared is equal to
x squared plus y squared plus z squared.
On the right side we have two z.
The equation is now in rectangular form
because it only contains the variables, x, y, and z,
but because we know it's a sphere,
let's write the equation in standard form.
Let's move the two z to the left side
which gives us x squared plus y squared plus z squared,
and then minus two z equals zero.
Let's complete the square on the z part of the equation,
so that the equation is in standard form.
This will allow us to find the center as well as the radius.
In order to complete the square on z squared minus two z,
we need to add the square of 1/2 of negative two,
the square of 1/2 of negative two
is the square of negative one over one,
so we add one to both sides of the equation
to complete the square on the z part.
This gives us x squared plus y squared,
and now to factor z squared minus two z plus one,
we need to find the factors of positive one
to add to negative two,
which are negative one and negative one,
which gives us two factors of z minus one,
or the quantity z minus one squared.
And this is equal to one.
So this is a rectangular equation for the sphere,
but here the equation is also in standard form,
which tells us that the center is zero comma zero comma one,
and the radius of the sphere is one.
I hope you found this helpful.