# Polar to Cartesian.

#### 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?

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#### matheagle

MHF Hall of Honor
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.

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#### phantomfn8

So is it possible to convert to cartesian form?

#### SpringFan25

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

MHF Helper
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
$$\displaystyle sin(\theta+ \phi)= sin(\theta)cos(\phi)+ cos(\theta)sin(\phi)$$\
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 $$\displaystyle r sin(\theta)= y$$.

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

#### phantomfn8

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?

#### phantomfn8

$$\displaystyle sin(\theta+ \phi)= sin(\theta)cos(\phi)+ cos(\theta)sin(\phi)$$\
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?