Robb Johns
Nov 12, 2024
Find the area enclosed by the curve x = t^2 – 2t , y=sqrt(t), and the y-axis?
What is the area enclosed by the curve defined by the equations ( x = t^2 - 2t ) and ( y = \sqrt{t} ), along with the y-axis? Please provide a detailed explanation of the steps involved in finding this area.
1 Answers
Dec 13, 2024
First find the the value of t where the curve intersects the Y-axis. This is when x = 0.
x = t^2 – 2t = 0 = t(t – 2)
So t= 0 and t = 2
dA = (0 – x)*dy …. Since the curve has negative x in this region
y = SQRT(t) and dy = [(1/2)/SQRT(t)]dt
dA = [2t – t^2][(1/2)/SQRT(t)]dt
dA = [t^(1/2) – (1/2)t^(3/2)]dt
Integrate to get: 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 = SQRT(2)[4/3 – 4/5]
Area = SQRT(2)[8/15) = 0.754
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