Area surface of revolution (rotating this astroid curve around the x-axis)

[SomeBody]

Homework Statement

Calculate the area of the surface of revolution obtained by rotating the following astroid curve around the x-axis:

Relevant Equations

x=a cos³(t),
y= a sin³(t)

Hello,

I am studying arc lengths and areas for parametric curves from the Adams & Essex Calculus book and I am a bit baffled by example 2 in the image attached. I understand the solution in the book where they integrate from t=0 to t=pi/2 (first quadrant) and multiply by two to get the full area.

However I tried to solve the exercise by integrating from t=0 to t=pi and get 0 as a result. How is this possible that the first and second quadrant cancel each other out by taking the integral over both?
- the arclength is positive for the whole interval t=0 to t=pi
- dt is positive for the whole interval t=0 to t=pi
- y = a sin³(t) is positive for the whole interval t=0 to t=pi
So the integral for the second quadrant is an infinite sum of positive terms but the integral from t=p/2 to t=pi is the negative opposite from the first quadrant. How is that possible?

 

[fresh_42]

I am not sure exactly what you calculated, but integrals are oriented volumes. It can easily happen that one part is right-handed and the other one left-handed, resulting in zero.

 

[Orodruin]

If you extend to $t \in (0,\pi)$ then $\sqrt{\cos^2 t} = |\cos t|$, which is not equal to $\cos t$ for $t > \pi/2$.

The cosine becomes negative.

Edit: So yes, the arc length should be positive, but the expression in the example is not. The author has used that $t < \pi/2$ to move the cosine outside the square root without the absolute value.