Finding the Lateral Surface Area with Line Integral
So my textbook has a challenge problem where it asks to find the surface area of a figure using line integrals. However, I'm not quite sure how it would be accomplished. The problem goes:
The figure below shows a piece of tin that has been cut from a circular cylinder. The base of the circular cylinder is modeled by$x^2 +y^2 = 9$. At any point $(x, y)$ on the base, the height of the object is
$$f(x,y) = 1 + \cos\left(\frac{\pi\cdot x}{4}\right).$$
Explain how to use a line integral to find the surface area of the piece of tin.
I considered converting everything into a parametric equation with variable $t$, and continuing from there. However, I don't think that method is right because then I would end up with the following:
$$f(x(t),y(t)) = 1 + \cos\left(\frac{\pi\cdot x(t)}{4}\right),$$
where $x(t) = 3\cos(t)$, which would result in a function that isn't elementary.
Any ideas?
Edit: In case you would like to see the actual problem out of the book, here is a link to it:
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$\begingroup$Use $r(t) = \langle 3\cos(t),3\sin(t)\rangle$ and then plug $x(t)=3\cos(t)$ into your $f(x,y)$ equation.
The lateral surface area is the line integral over your given curve, so it would be integral of $f(x,y)\,\mathrm ds$ where $\mathrm ds = || r'(t)||\, dt = 3\, dt$, and $t$ would go from $0$ to $2\pi,$ since it acts like $\theta$.
Hope that helps.
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