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

Solution:

First, find the value of t where the curve intersects the y - axis i.e. x = 0

Given: x = t² - 2t

⇒ t² - 2t = 0

⇒ t(t - 2) = 0

So t = 0 and t = 2.

Derivative of area, dA/dy

dA/dy = (0 - x)

dA = (0 - x) × dy

Since the curve has negative x in this region.

y = √t (given)

dy/dt = (1/2)/√t

dA = [0 - (t² - 2t)] [(1/2)/√t]dt

dA = [2t - t²]/[(1/2)/√t]dt

dA = [t^1/2 - (1/2)t^3/2]dt

Integrating to obtain A,

A = (2/3)t^3/2 - (1/5)t^5/2

Now evaluate from t = 0 to t = 2

Area = [(2/3)2^3/2 - (1/5)2^5/2] - [0]

Area = √[2(4/3 - 4/5)]

= √[2(8/15)]

= √(16/15)

= 0.754

Therefore, the area enclosed by the curve is 0.754 square units.

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

Summary:

The area enclosed by the curve x = t² - 2t , y = √(t), and the y-axis is 0.754 square units.