# 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!

## 3 Answers

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.

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 $$x=33$$

• I understand what you've done and I've tried doing it this way. However, I've failed to come up with the correct answer (I got -273.0666). Thank you for trying though! – CodingMee Mar 20 at 3:13
• I believe the answer should be positive, but that is the answer Alpha gets. Why do you think it is wrong? – Ross Millikan Mar 20 at 3:35
• I tried both negative and positive just to be sure, and they both come up incorrect. I think it is wrong because my assignment is online and automatically checks our answers. – CodingMee Mar 20 at 3:40
• The answer key could be wrong, it could be expecting the exact answer of $\frac {4096}{15}$, or I could be interpreting the problem wrong. – Ross Millikan Mar 20 at 3:44

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 $$\frac{4096}{15}\approx 273.067$$