You mean, the area surrounded by the 3 curves. Not surrounding the curves.

a) y1 = x, y2 = 1/sqrt(x), x = 2

Sketching the graphs of those 3, the easiest area is by making the

dA = (y1 -y2)dx

The integration, or the boundaries of dx, then is from x=1 to x=2.

x=1 being at the intersection of curves y1 and y2.

So,

A = INT.(1-->2)[x - 1/sqrt(x)]dx

A = INT.(1-->2)[x -x^(-1/2)]dx.

A = [(1/2)x^2 +2sqrt(x)]|(1-->2)

A = [(1/2)(2^2) +2sqrt(2) -0 -0]

A = 2 +2sqrt(2) = 2(1 +sqrt(2)) = 4.82843 sq.units --------answer.

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b) sqrt(x) + sqrt(y) = 1, x = 0, y = 0

I don't know how to graph the "sqrt(x) + sqrt(y) = 1" now.

If you know how to use a graphing calculator, see how goes the graph of that. Then it should be easy to choose your dA.