Area Inside A Loop Formed By Parametric Equations We are given:
$x=49-t^2$
$y=t^3-16t$
The curve apparently makes a loop which lies along the x-axis. I need help finding total area inside the loop. I don't know where to even start.
If it helps, in previous parts of the question, I found that 
(a) the tangent line is horizontal at $t=\sqrt{16/3}$ and $x = 43.6666666666667$
(b) the tangent line is vertical at $t=0$
Thank you!
 A: I think this is a Green's theorem problem.  The loop is traced as $t$ goes from $-4$ to $4$ (but clockwise.)  So the area is
$$-\frac{1}{2} \int_{-4}^4 x \; dy - y \; dx $$ $$ =
-\frac{1}{2} \int_{-4}^{4} (49-t^2)(3t^2-16)-(t^3-16t)(-2t) \; dt $$ $$= \frac{8192}{15} = 546.13\ldots.$$
The extra minus sign is because of the clockwise orientation.
A: Have you plotted it?  I don't find a loop.  I find an arc below the $x$ axis from $t=0$ to $t=4$.    This is from $x=33$ to $x=49$.  Maybe you are supposed to find the area between the $x$ axis and the curve in this region.  Here is my plot

$t=0$ is the right hand end, at $(49,0).\ \ t=4$ is the point where it crosses the axis at $(33,0)$.  If you are to find the area between the curve and the $x$ axis you can just solve the $x$ equation for $t$ and plug into the $y$ equation
$$x-49-t^2\\t=\sqrt{49-x}\\y=(49-x)^{3/2}-16(49-x)^{1/2}$$
and you can integrate the last from $x=49$ to $x=33$
A: Making a parametric plot, there is effectively a loop which is symmetric with respect to the $x$ axis; the points where the curve intersect the axis correspond to $x=33$ and $x=49$ as @Ross Millikan already answered.
The major issue is to compute
$$I=\int \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt=\int\sqrt{4 t^2+\left(3 t^2-16\right)^2}\,dt$$ which would lead to nasty elliptic integrals.
So, the simplest is to do what @Ross Millikan already answered, that is to say
$$t=\pm \sqrt{49-x} \implies y=\pm (x-33)\sqrt{49-x} $$ So, the total area enclosed by the loop is 
$$A=2\int_{33}^{49}(x-33)\sqrt{49-x}\,dx$$ Using
$$\int(x-33)\sqrt{49-x}\,dx=-\frac{2}{15} (49-x)^{3/2} (3 x-67)$$ we end with $A=2 \times \frac{4096}{15}$.
Notice that $\frac{4096}{15}\approx 273.067$
