Question 1036835
<pre><b>

{{{r}}}{{{""=""}}}{{{6sin(theta)}}}

If we draw the graph of that on polar graph paper. we make
a table of points:

  <font face="symbol">q</font> | r
  0院0.00
 15院1.55
 30院3.00
 45院4.24
 60院5.20
 75院5.80
 90院6.00
105院5.80
120院5.20
135院4.24
150院3.00
165院1.55
180院0.00

We get this circle:

{{{drawing(400,400,-8,8,-8,8,

green(line(15,0,-15,0),
line(14.4888874,3.88228568,-14.4888874,-3.88228568),
line(12.9903811,7.5,-12.9903811,-7.5),
line(10.6066017,10.6066017,-10.6066017,-10.6066017),
line(7.5,12.9903811,-7.5,-12.9903811),
line(3.88228568,14.4888874,-3.88228568,-14.4888874),
line(0,15,0,-15),
line(-3.88228568,14.4888874,3.88228568,-14.4888874),
line(-7.5,12.9903811,7.5,-12.9903811),
line(-10.6066017,10.6066017,10.6066017,-10.6066017),
line(-12.9903811,7.5,12.9903811,-7.5),
line(-14.4888874,3.88228568,14.4888874,-3.88228568),
line(-15,0,15,0),
line(-14.4888874,-3.88228568,14.4888874,3.88228568),
line(-12.9903811,-7.5,12.9903811,7.5),
line(-10.6066017,-10.6066017,10.6066017,10.6066017),
line(-7.5,-12.9903811,7.5,12.9903811),
line(-3.88228568,-14.4888874,3.88228568,14.4888874),
line(0,-15,0,15),
line(3.88228568,-14.4888874,-3.88228568,14.4888874),
line(7.5,-12.9903811,-7.5,12.9903811),
line(10.6066017,-10.6066017,-10.6066017,10.6066017),
line(12.9903811,-7.5,-12.9903811,7.5),
line(14.4888874,-3.88228568,-14.4888874,3.88228568),
line(15,0,-15,0),
circle(0,0,1), circle(0,0,2),circle(0,0,3), circle(0,0,4),
circle(0,0,5),circle(0,0,6),circle(0,0,7), circle(0,0,8),
circle(0,0,9),circle(0,0,10)),

circle(0,3,3)


 )}}}

When changing from one form to the other always draw this first:

{{{drawing(200,100,-.5,2,-.5,1.5,
line(-3,0,3,0),line(0,-3,0,3), line(0,0,1.5,1),line(1.5,0,1.5,1),
locate(.4,.3,theta), locate(1.53,.62,y),locate(.77,0.05,x),
locate(.75,.85,r) )}}}

We substitute for the sine first.
Looking at the triangle and remembering that
{{{SINE=OPPOSITE/HYPOTENUSE}}} 

{{{sin(theta)=y/r}}}

so substitute {{{y/r}}} for {{{sin(theta)}}} first

{{{r}}}{{{""=""}}}{{{6(y/r)}}}

{{{r}}}{{{""=""}}}{{{(6y)/r}}}

Multiply both sides by r

{{{r^2}}}{{{""=""}}}{{{6y}}}

Now use the Pythagorean theorem to substitute {{{x^2+y^2}}}
for {{{r^2}}}

{{{x^2+y^2}}}{{{""=""}}}{{{6y}}}  <-- answer

As a check let's graph it.

Get it in standard form:

{{{x^2+y^2-6y}}}{{{""=""}}}{{{0}}}

Write x as (x-0)

Complethe square:
take half or -6, get -3
Square it, get 9
Add 9 to both sides:

{{{(x-0)^2+y^2-6y+9}}}{{{""=""}}}{{{9}}}

Factor the trinomial in y:

{{{(x-0)^2+(y-3)^2}}}{{{""=""}}}{{{9}}}

That's a circle with center (0,3) and radius 3:

{{{drawing(400,400,-8,8,-8,8,graph(400,400,-8,8,-8,8),
circle(0,3,3))}}}

It graphs as the same circle, so we are right.

Edwin</pre></b>