1. ## Polar to Cartesian.

In my calculus class, I chose to doa project on finding the area inside a polar curve and then revolving the curve around the x-axis in order to find the surface area and the volume of the solid. Unfortunately, I got stuck on converting r=2.5sin(3θ) into Cartesian form. Is this possible and if so how is it done?

2. Originally Posted by phantomfn8
In my calculus class, I chose to doa project on finding the area inside a polar curve and then revolving the curve around the x-axis in order to find the surface area and the volume of the solid. Unfortunately, I got stuck on converting r=2.5sin(3θ) into Cartesian form. Is this possible and if so how is it done?

I thought it was just another circle.
I believe it's a 3 leaf petal now.

3. So is it possible to convert to cartesian form?

4. You can either convert it because you know the cartesian equation of the circle matheagle described, or you can use the formula below
$x=rcos(\theta)$
$y=rsin(\theta)$

These can be used to convert any polar coordinates $(r,\theta)$ to cartesian coordinates $x,y$, but it looks messy....id work out the equation of the circle instead.

5. Originally Posted by phantomfn8
In my calculus class, I chose to doa project on finding the area inside a polar curve and then revolving the curve around the x-axis in order to find the surface area and the volume of the solid. Unfortunately, I got stuck on converting r=2.5sin(3θ) into Cartesian form. Is this possible and if so how is it done?
a single "petal", or is the whole thing to be rotated about the x-axis?

for the volume, you might consider Pappus's Centroid Theorem -- from Wolfram MathWorld

6. $sin(\theta+ \phi)= sin(\theta)cos(\phi)+ cos(\theta)sin(\phi)$\
and
$cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(\theta)sin(\phi)$.

In particular,
$sin(2\theta)= sin(\theta+ \theta)$ $= sin(\theta)cos(\theta)+ cos(\theta)sin(\theta)= 2sin(\theta)cos(\theta)$
and
$cos(2\theta)= cos(\theta+ \theta)$ $= cos(\theta)cos(\theta)- sin(\theta)sin(\theta)= cos^2(\theta)- sin^2(\theta)$

Now,
$sin(3\theta)= sin(2\theta+ \theta)= sin(2\theta)cos(\theta)+ cos(2\theta)sin(\theta)$ $= 2sin(\theta)cos^2(\theta)+ (cos^2(\theta)- sin^2(\theta))sin(\theta)$ $= 2sin(\theta)cos^2(\theta)+ cos^2(\theta)sin(\theta)- sin^3(\theta)$ $= 3sin(\theta)cos^2(\theta)- sin^3(\theta)$.

So $r= .5 sin(3\theta)= .5(3 sin(\theta)cos^2(\theta)- sin^3(\theta)$

Multiply both sides by $r^3$ to get
$r^4= 1.5 (r sin(\theta)(r^2 cos^2(\theta))- .5 (r^3 sin^2(\theta))$

$r^2= x^2+ y^2$ so $r^4= (x^2+ y^2)^2$ while $r cos(\theta)= x$ and $r sin(\theta)= y$.

$(x^2+ y^2)^2= 1.5 xy^2- .5 y^3$

7. I intended odor the entire curve to be rotated around the x-axis. How do you find the volume of parametric equations rotated around the axis?

8. Originally Posted by HallsofIvy
$sin(\theta+ \phi)= sin(\theta)cos(\phi)+ cos(\theta)sin(\phi)$\
and
$cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(\theta)sin(\phi)$.

In particular,
$sin(2\theta)= sin(\theta+ \theta)$ $= sin(\theta)cos(\theta)+ cos(\theta)sin(\theta)= 2sin(\theta)cos(\theta)$
and
$cos(2\theta)= cos(\theta+ \theta)$ $= cos(\theta)cos(\theta)- sin(\theta)sin(\theta)= cos^2(\theta)- sin^2(\theta)$

Now,
$sin(3\theta)= sin(2\theta+ \theta)= sin(2\theta)cos(\theta)+ cos(2\theta)sin(\theta)$ $= 2sin(\theta)cos^2(\theta)+ (cos^2(\theta)- sin^2(\theta))sin(\theta)$ $= 2sin(\theta)cos^2(\theta)+ cos^2(\theta)sin(\theta)- sin^3(\theta)$ $= 3sin(\theta)cos^2(\theta)- sin^3(\theta)$.

So $r= .5 sin(3\theta)= .5(3 sin(\theta)cos^2(\theta)- sin^3(\theta)$

Multiply both sides by $r^3$ to get
$r^4= 1.5 (r sin(\theta)(r^2 cos^2(\theta))- .5 (r^3 sin^2(\theta))$

$r^2= x^2+ y^2$ so $r^4= (x^2+ y^2)^2$ while $r cos(\theta)= x$ and $r sin(\theta)= y$.

$(x^2+ y^2)^2= 1.5 xy^2- .5 y^3$
Thank you soo much! Now would ou know how to find the volume of the graph when rotate around the x-axis?