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r = 1 + sin(4*Θ)

where 0<= Θ <= 2pi

I understand that I can find surface area of parametric equations using

S = ʃ 2(pi)y sqrt([dx/dt]² + [dy/dt]²) dt

(a->b)

I'm also familiar with:

x = r cos(Θ)

y = r sin(Θ)

r = sqrt(x² + y²)

Θ = tan(y/x)

And lastly, with all my efforts, basically all I did was write a bunch of stuff down and hope something jumped out at me.

I drew a triangle and labeled it...(you can laugh at my attempt to draw it, you can also tell where theta is supposed to go)

.

|\

y | \ r = 1 + sin(4Θ) = sqrt(x² + y²)

| \

----`

x

and so sin(Θ) = y / (1 + sin(4Θ))

and sin(Θ) = y / (sqrt(x² + y²))

1 + sin(4Θ) = sqrt(x² + y²)

but it turns out I don't really know what I'm doing, I have no direction. Any help is greatly appreciated, thanks a lot!