How Do You Calculate the Surface Area of a Cardioid Revolved Around the X-Axis?

[iqjump123]

Homework Statement

Find the area of the surface of revolution generated by revolving about the x-axis the cardioid x=2cos[itex]\vartheta[/itex]-cos2[itex]\vartheta[/itex],y=2sin[itex]\vartheta[/itex]-sin2[itex]\vartheta[/itex].

pictures show a cardioid with [itex]\vartheta[/itex]=pi and [itex]\vartheta[/itex]=0.

Homework Equations

After looking up, the formula to solving this type of equation is:
s=int(sqrt(dx/dtheta)^2 +sqrt(dy/dtheta)^2)
and also some trig identities

The Attempt at a Solution

Alright, I first went ahead and applied the equation above to the functions, and integrated from 0 to pi (since that is where the edges of the cartoid lie?). However, after going through the math and solving it out, I get a -2/3(sinx*cosx^3/2) from 0 to pi, which makes no sense (since volume will be 0 if this int. is evaluated).

thanks so much in advance!

 

[mighty2000]

Thank you for your post. Finding the area of the surface of revolution generated by revolving a curve about an axis is a common problem in mathematics and physics. In this case, the curve is a cardioid, which is a special type of curve often seen in nature.

To find the area of the surface of revolution, we can use the formula:

A = 2π∫y√(1+(dy/dx)^2)dx

In this case, we are revolving the curve about the x-axis, so we will use the formula:

A = 2π∫y√(1+(dy/dx)^2)dx

where y is the function of x that represents the cardioid curve. In this case, we have x=2cosθ-cos2θ and y=2sinθ-sin2θ.

To use the formula, we need to find the derivative of y with respect to x, which can be done using the chain rule:

dy/dx = (dy/dθ)(dθ/dx)

Using the trig identities, we can simplify this to:

dy/dx = 2cosθ-sinθ

Substituting this into the formula, we get:

A = 2π∫(2sinθ-sin2θ)√(1+(2cosθ-sinθ)^2)dθ

Now, we can use trigonometric identities to simplify this integral. After some algebraic manipulation, we get:

A = 2π∫(2sinθ-sin2θ)√(5-4sinθ)dθ

We can now use the substitution u = 5-4sinθ to simplify the integral further:

A = 2π∫(2sinθ-sin2θ)√u(du/-4cosθ)

= 2π∫(sinθ-sin2θ)√u(-du/2cosθ)

= -π∫(sin2θ-sin3θ)√u(du/2cosθ)

= -π∫(sin2θ-sin3θ)√(5-4sinθ)(du/2cosθ)

= -π∫(sin2θ-sin3θ)√(5-4sinθ)(du/2cosθ)

= -π∫(sin2θ-sin3θ