1. ## Parametric Loop Area

Notice that the curve given by the parametric equations

The curve makes a loop which lies along the x-axis. What is the total area inside the loop?

What integral do I need to integrate, or what better method can I use, to find this? Please help!

2. Originally Posted by Del
Notice that the curve given by the parametric equations

The curve makes a loop which lies along the x-axis. What is the total area inside the loop?

What integral do I need to integrate, or what better method can I use, to find this? Please help!
Always draw a graph it really helps

As you noted the graph is symmetric with respect to the x axis. So we can find the area above the x axis and double it.

$\displaystyle y=t^3-25=t(t-5)(t+5)$
so the y intercepts are at t= -5,0,5

$\displaystyle \int_{-5}^{0}ydx=\int_{-5}^{0}(t^3-25t)(-2tdt)=\int_{-5}^{0}-2t^4+50t^2dt$

$\displaystyle \int_{-5}^{0}-2t^4+50t^2dt=-\frac{2}{5}t^5+\frac{50}{3}t^3|_{-5}^{0}= \frac{2}{5}(0)^5+\frac{50}{3}(0)^3- \left( \frac{2}{5}(-5)^5+\frac{50}{3}(-5)^3 \right)=\frac{10000}{3}$

This is half of the area(The part above the x axis)

So multipling by 2 we get $\displaystyle \frac{20000}{3}$

3. That makes sense all the way, but the program I am inputting my answers into is saying that this is wrong. Thanks though!

4. Pesky minus sign Dang. I still can't add

$\displaystyle \int_{-5}^{0}-2t^4+50t^2dt=-\frac{2}{5}t^5+\frac{50}{3}t^3|_{-5}^{0}= \frac{2}{5}(0)^5+\frac{50}{3}(0)^3- \left( \frac{2}{5}(-5)^5+\frac{50}{3}(-5)^3 \right)=\frac{10000}{3}$

Notice the missing negative after the large parenthsis before the 2/5.

$\displaystyle -\frac{2}{5}t^5+\frac{50}{3}t^3|_{-5}^{0}= -\frac{2}{5}(0)^5+\frac{50}{3}(0)^3- \left(- \frac{2}{5}(-5)^5+\frac{50}{3}(-5)^3 \right)=\frac{2500}{3}$

mult by 2 to get $\displaystyle \frac{5000}{3}$

5. Thanks a ton, you really cleared this up!