parametric equations area?

[sphynx_000]

the curve is x=10-t^2, y=t^3-12t

when I use the equation A=integral (y dx), im finding area using columns right?
Then I look at the x axis (limits form x=-2 to x=10 changed to t=12^.5 to t=0)
will that find me the entire area, or just half>>????




Im confused could anyone help me out?

 

[sphynx_000]

wait,....


that would be +-12^.5 ..... so im only finding the top or bottom half of the curve.



I think i figured it out, someone correct me if im wrong!

 

[galactus]

\(\displaystyle \begin{array}x=10-t^{2}...[1]\\y=t^{3}-12t...[2]\end{array}\)

Solve [1] for x and sub into [2] and you get:

\(\displaystyle \L\\(10-x)^{\frac{3}{2}}-12\sqrt{10-x} \;\ and \;\ 12\sqrt{10-x}-(10-x)^{\frac{3}{2}}\)

Therefore, as you suspected, multiply by 2:

\(\displaystyle \L\\2\int_{-2}^{10}\left[12\sqrt{10-x}-(10-x)^{\frac{3}{2}}\right]dx=2\int_{-2}^{10}\left[x\sqrt{10-x}+2\sqrt{10-x}\right]dx\)

 

[sphynx_000]

thank you!