Area under a Parametric Curve

[Kyle]

[math]x=81-t^2[/math]

[math]y=t^3-16t[/math]

After asking about tangent lines, the question asks "The curve makes a loop which lies along the x-axis. What is the total area inside the loop?"

I thought I was supposed to combine these equations so that there were no t's and then integrate it.

I found that [math]y=(81-x)^{3/2}-16\sqrt{81-x}[/math]

And, after looking at the curve, it looks like the answer for the Area should be

[math]\int_{65}^{81} [(81-x)^{3/2}-16\sqrt{81-x}] \,dx=-4096/15[/math]

I've tried this answer both negative and positive and it's been wrong. Maybe I haven't found the equation of the curve correctly or integrated correctly. Does anyone see my mistake?

[EvoN1020v]

Upon graphing those two functions together, I don't see any loop?

[insane_alien]

i also don't see a loop.

[the tree]

Upon graphing those two functions together, I don't see any loop?

Isn't that one function, expressed parametrically? Anyways, I'll grab my textbook and have a proper answer for this in a bit...

 

edit, got it: to find the area under a curve expressed parametrically you use [math]\int_a^b y \frac{dx}{dt} \cdot dt[/math]

[Kyle]

edit, got it: to find the area under a curve expressed parametrically you use [math]\int_a^b y \frac{dx}{dt} \cdot dt[/math]

[math]

\int_{65}^{81} [(81-x)^{3/2}-16\sqrt{81-x}] \,dx=-4096/15

[/math]

[math]

\int_{0}^{4} (t^3-16t)(-2t) \,dt=4096/15

[/math]

I wonder if there might be a technical problem with this question. I think I have the right answer and your equation confirms it. I'll ask a few more people around the dorms and see if anyone has figured out what my teacher was trying to ask.