Since you are talking about limits, I am assuming you are wanting to calculate the area using a Riemann Sum.

I will not take your hint. The area between x = 2y - y^2 and the y-axis and the area between y = 2x - x^2 and the x-axis are exactly the same, so you can just compute the area between x = 2y - y^2 and the y-axis just like you would when you compute the area between y = 2x - x^2.

Note that x = 2y - y^2 intersects the y-axis when:
2y - y^2 = 0 ==> y = 0 and y = 2.

Then, split the region into n rectangles. The width of each rectangle is 2/n (the width of the interval [0, 2] divided by the number of rectangles). Next, the height of the i-th rectangle, where i = 1, 2, ..., n, is just the height of the function, x = 2y - y^2.

Since the i-th rectangle has it's bottom-left corner at (0, 2i/n), we see that the area of the i-th rectangle is:
A(i-th rectangle) = bh = (2/n)[2(2i/n) - (2i/n)^2] = 8i/n^2 - 8i^2/n^3.

The sum of n of these rectangles is:
sum(i=1 to n) (8i/n^2 - 8i^2/n^3)
= (4/n^2)sum(i=1 to n) i - (8/n^3)sum(i=1 to n) i^2
= (4/n^2)[n(n + 1)/2] - (8/n^3)][n(n + 1)(2n + 1)/6]
= 2n(n + 1)/n^2 - (4/3)n(n + 1)(2n + 1)/n^3

Letting n --> infinity yields the area to be 2 - (4/3)(2) = 4/3.

I hope this helps!