Calculus applications of integration problem?

Question

Find the area bounded by the indicated curve:

y = (2) / (sqrt X)
X = 0, y = 1 , y = 3

Answer 1

I'd start by rewriting this as a function of y.

y = 2 / sqrt(x)
sqrt(x) = 2 / y
x = 4 / y^2 = 4y^-2

Now we simply need to integrate this with respect do dy with bounds of 1 and 3.

The integral of 4y^-2 dy from 1 to 3 is -4y^-1 = -4 / y evaluated from 1 to 3, which is (-4 / 3) - (-4 / 1) = -4/3 + 4 = 8/3.

Answer 2

The limits of integration are given in the y coordinates.
What is x as a function of y?

y = 2 / sqrt(x)
sqrt x = 2 / y
x = 4 / y^2

Now integrate wrt y.

A = Integral(low 1 up 3) 4 / y^2 dy
A = 4 Integral(low 1 up 3) y^-2 dy
A = -4 Eval(low 1 up 3) y^-1
A = -4 (1/3 - 1)
A = -4 (-2/3)
A = 8/3

Answer 3

This curve could nicely be restated as a functionality of y. y = 2 / sqrt(x) y/2 = a million / sqrt(x) a million / x = y^2 / 4 x = 4 / y^2 = 4y^(-2) necessary from a million to 3 of 4y^(-2)dy = 4/(-a million) * y^(-a million) (from a million to 3) = -4 / y (from a million to 3) = (-4/3 - (-4/a million)) = 8/3 the respond is 8/3, no longer 3.