# What is the Integral for the Area of a Region Enclosed by a Parametric Equation?

• PolarClaw
In summary, the conversation discusses setting up an integral for the area of a region enclosed by the graph x=t^2-2 and y=t^3-2t. The parametrization is broken into three implicit functions, y = x\sqrt{x+2} for t\in(0, \infty), y = -x\sqrt{x+2} for t\in(-\infty, 0), and y = 0 for t = 0, which enclose a region between x=0 and x=-2. The integral can be evaluated by integrating the positive function on this interval and multiplying by 2 due to the symmetry of the graphs.
PolarClaw
Hi,

I've been trying to do this one question: Let R be the region enclosed by the graph x=t^2-2 y=t^3-2t. Set up the integral for the area of R.

I know that if y is continuous function of x on an interval a ≤ x ≤ b where x=f(t) and y=g(t) then $$\int_{a}^{b}$$ y dx =$$\int_{t1}^{t2}$$ g(t)f'(t)dt provided that f(t1)=a and f(t2)=b and both g and f' are continuous on [t1,t2].

But everytime i equate and solve for t i get sqrt(2) and 1. This is where i get lost I can't seem to find the interval for parametric equations (any of them ). Other then the interval I do believe i have the rest of the integral:

$$\int$$ (t^3-2t)(2t)dt

Thanks in advance (sorry for long post as well as horrid use of that latex typing first time)

PolarClaw said:
Hi,

I've been trying to do this one question: Let R be the region enclosed by the graph x=t^2-2 y=t^3-2t. Set up the integral for the area of R.

The graph translated to $$y=x\sqrt{x+2}$$. There doesn't seem to be any enclosed region.

the question copied and pasted from the pdf file is as follows:

4. Let R be the region enclosed by the graph of x = t^2 − 2, y = t^3 − 2t . (The graph was sketch for the final in 2004)

a. Set up, BUT DO NOT EVALUATE an integral for the area of R .

Thats all the info that was given :/

Last edited:
Sorry, there is NO area "enclosed" by that graph.

Actually, the parametrization can be broken into the 3 implicit functions y = $x\sqrt{x+2}$ for $t\in(0, \infty)$, y = $-x\sqrt{x+2}$ for $t\in(-\infty, 0)$, and y = 0 for t = 0. The graphs of these functions enclose a region between x=0 and x=-2. Given the symmetry of the graphs of the functions (one is just a reflection in the x-axis of the other), all one needs to do is integrate the positive function on this interval and multiply by 2.

Thanks for the help got was able to get the answer now.

hypermorphism said:
Actually, the parametrization can be broken into the 3 implicit functions y = $x\sqrt{x+2}$ for $t\in(0, \infty)$, y = $-x\sqrt{x+2}$ for $t\in(-\infty, 0)$, and y = 0 for t = 0. The graphs of these functions enclose a region between x=0 and x=-2. Given the symmetry of the graphs of the functions (one is just a reflection in the x-axis of the other), all one needs to do is integrate the positive function on this interval and multiply by 2.

Oops. Yes... the minus sign. Sorry about that.

## What is the definition of area in parametric equations?

The area in parametric equations refers to the region enclosed by a parametric curve on a coordinate plane.

## How is the area of a parametric curve calculated?

The area of a parametric curve can be calculated by using the formula A = ∫(y(t)x'(t))dt, where y(t) and x'(t) represent the y and x coordinates of the curve, respectively.

## Can the area of a parametric curve be negative?

No, the area of a parametric curve cannot be negative as it represents a physical region and cannot have a negative value.

## What are the limitations of using parametric equations to calculate area?

Parametric equations can only be used to calculate the area of simple curves, such as circles, ellipses, and parabolas. They cannot be used for more complex shapes or irregular curves.

## How is the area of a parametric curve affected by the parameterization of the curve?

The area of a parametric curve is not affected by the parameterization of the curve, as long as the curve is traced in the same direction. However, the method of calculation may vary depending on the parameterization used.

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