Question

Find the area enclosed by the curve \(x = {t^2} - 2t,y = \sqrt t \) and the y-axis.

Short answer

The area enclosed of the given curve is given as: \(A = \frac{{8\sqrt 2 }}{{15}}\)

Step-by-step solution

Draw the graph of the given expression

The parametric equation of an curve is:

\(\begin{array}{l}x = {t^2} - 2t\\y = \sqrt t \end{array}\)

The graph of the ellipse when is shown below:

Find the area

The area enclosed by the curve\(x = f(t),y = g(t)\)as \(t\)increases from\(\alpha \)to\(\beta \)is:

\(A = \int\limits_\alpha ^\beta {g(t)f'(t)} dt\)

Here\(f(t) = {t^2} - 2t,g(t) = \sqrt t \)

Find\(f'(t)\)

\(\begin{array}{l}f(t) = {t^2} - 2tf\\f'(t) = 2t - 2\end{array}\)

Now we need to find the interval of t.

Since the curve has intersections with y-axis so we need to let x equal to zero to get the values of t.

\(\begin{array}{l}{t^2} - 2t = 0\\t(t - 2) = 0\\t = 0,2\end{array}\)

The interval will be: \(\left( {0,2} \right)\)

The area of the curve is:

\(\begin{array}{l}A = \int\limits_\alpha ^\beta {g(t)f'(t)} dt\\A = \int\limits_0^2 {\sqrt t (2t - 2)} dt\\A = \int\limits_0^2 {(2{t^{\frac{3}{2}}} - 2\sqrt t )} dt\\A = 2\left( {\frac{2}{5}{t^{\frac{5}{2}}}} \right)_0^2 - 2\left( {\frac{2}{3}{t^{\frac{3}{2}}}} \right)_0^2\\A = \frac{{8\sqrt 2 }}{{15}}\end{array}\)

Hence, the area enclosed of the given curve is given as: \(A = \frac{{8\sqrt 2 }}{{15}}\).