Area under a parabolic trajectory

Question

I have this problem:

"prove that the area under the trajectory described by a parabolic shot that has:

$f(x)=\tan(\theta)x - (\frac{g}{2v^2\cos^2(\theta)})x^2$

and

$x=v\cos(\theta)t$

is defined as:

$A(\theta)=\frac{2v^4}{3g^2}\sin^3(\theta)\cos(\theta)$

I have tried to change x in order to convert $f(x)$ to a function of $\theta$ but i end up everytime with $f(\theta)=0$

Answer 1

This is a straight-forward integral. $\theta$ is fixed for the projectile. If $\theta=0$ then the problem does not make much sense. If $\theta=\pi/2$ then the area is undefined. So it makes sense to assume $\theta\in(0,\pi/2)\cup(\pi/2,\pi)$

$$f(x)=0 \iff x=0 \ \ \vee \ \ x=\dfrac{2v^2\cos^2\theta\tan\theta}{g}$$

And so,

$$A(\theta) = \int_0^{2g^{-1}v^2\cos^2\theta\tan\theta} x\tan\theta - \frac{g}{2v^2\cos^2\theta}x^2 \ dx$$

which is a simple integral (polynomial in $x$).

Addendum:

It is well known, and relatively easy to prove using calculus, that the area under a parabolic arch is $A=\dfrac{2}{3}bh$ where $b$ is the base of the arch and $h$ its height.

The height of a parabolic arch $f(x)=ax^2+bx+c$ where $a,b,c$ make sense is $f(t)$ where $t=-\dfrac{b}{2a}$. In your case $t = \dfrac{v^2\cos^2\theta\tan\theta}{g}$, so

$$ h=\tan\theta\left(\dfrac{v^2\cos^2\theta\tan\theta}{g}\right) - \dfrac{g}{2v^2\cos^2\theta}\left(\dfrac{v^2\cos^2\theta\tan\theta}{g}\right)^2$$

And the base $b$ is the distance between the roots,

$$b=\dfrac{2v^2\cos^2\theta\tan\theta}{g}$$

This could be a useful exercise in algebra/identities if nothing else.