# Use double integral to find the area enclosed between y^2 = x^3 and y = x.

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As you can see on the graph the area in question is enclosed between the positive branch of `y^2 = x^3` (`y = x^(3/2)` ) and y = x

These two graphs intersect at (0,0) and (1,1)

As a double integral this can be written as

`int_0^1 int_(x^(3/2))^ xdydx`

If you take the inner intergral first (`int_(x^(3/2)) ^ x dy` ) this will reduce to

`int_0^1 (x - x^(3/2))dx`

which is what you would write if you were required to express the given area as a difference between the areas under the two curves.

This integrals equals to

`(x^2 / 2 - 2/5 * x^(5/2)) |_0 ^1 = 1/2 - 2/5 - (0 - 0) = 1/10`

**The area is 1/10.**