Polar to Cartesian.

May 2010
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0
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?
 
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matheagle

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Feb 2009
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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.
 
Last edited:
May 2010
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You can either convert it because you know the cartesian equation of the circle matheagle described, or you can use the formula below
\(\displaystyle x=rcos(\theta)\)
\(\displaystyle y=rsin(\theta)\)

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

skeeter

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Jun 2008
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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
 

HallsofIvy

MHF Helper
Apr 2005
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\(\displaystyle sin(\theta+ \phi)= sin(\theta)cos(\phi)+ cos(\theta)sin(\phi)\)\
and
\(\displaystyle cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(\theta)sin(\phi)\).

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

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

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

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

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

\(\displaystyle (x^2+ y^2)^2= 1.5 xy^2- .5 y^3\)
 
May 2010
4
0
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?
 
May 2010
4
0
\(\displaystyle sin(\theta+ \phi)= sin(\theta)cos(\phi)+ cos(\theta)sin(\phi)\)\
and
\(\displaystyle cos(\theta+ \phi)= cos(\theta)cos(\phi)- sin(\theta)sin(\phi)\).

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

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

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

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

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

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