Question

Find the area enclosed by the curve
x = t2 − 2t, y = t and the y-axis.

Answer 1

To find the enclosed area, identify the curve's intersection points with the y-axis by setting x equal to 0 to get t values. Set up an integral with these limits, substitute y and the differentiated x (dx), and then solve the integral.

Step-by-step explanation:

To find the area enclosed by the curve given by x = t^2 - 2t, y = t, and the y-axis, we can follow several steps. Since the curve is parametric, we cannot directly integrate it with respect to x or y without eliminating the parameter t; however, because we are also given the constraint of the y-axis, we can look for points where the curve intersects the y-axis (x=0) and use those as limits of integration.

To find the intersection points, we must set x equal to 0:

0 = t^2 - 2t

0 = t(t - 2)

This yields two solutions for t: t = 0 and t = 2. These are the points where the curve intersects the y-axis.

Next, we can set up the integral to find the area:

A = ∫_{t=0}^{t=2} y dx

To find dx, we differentiate x concerning t:

dx/dt = 2t - 2

dx = (2t - 2)dt

Substituting y = t and dx from above into the integral, we get:

A = ∫_{0}^{2} t(2t - 2)dt

This integral can be simplified and then solved using basic integration techniques.

Answer 2

The curve hits the y-axis whenever
`x=0`:


`x=t^2-2t=0\implies t(t-2)=0\implies t=0,t=2`

The area enclosed by the region is the sum of distances of every point on the curve where
`t\in[0,2]` to the origin. This is given by the integral


`\displaystyle\int_0^2√(x(t)^2+y(t)^2)\,\mathrm dt`

`\displaystyle\int_0^2√((t^2-2t)^2+t^2)\,\mathrm dt`

`\displaystyle\int_0^2√(t^2(t-2)^2)\,\mathrm dt`

`\displaystyle\int_0^2|t||t-2|\,\mathrm dt`

Since
`t\in[0,2]`, you have
`|t|=t` and
`|t-2|=-(t-2)=2-t`, giving you


`\displaystyle\int_0^2(2t-t^2)\,\mathrm dt=t^2-\frac{t^3}3\bigg|_(t=0)^(t=2)=4-\frac83=\frac43`