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I need to find the volume of the solid formed by the area trapped between $\ y=x^2$ and y=√x, rotated about the y-axis. The two curves cross at y = 0 and y = 1 so they will be the upper/lower limits.

Since we are rotating about the y-axis, I believe the y=√x curve is the inner curve and $\ y=x^2$ is the outer curve.

When I am finding the outer and inner radius do they become negative since I am taking the difference from the the y-axis (x=0)?

Right now I have

V= $∫[ π(x^2)^2 −π(√x)^2]dx = \frac{-3π}{10} $

I know something is wrong since the volume shouldn't be negative.

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The curve $y=x^2$ is indeed the outer curve, so in terms of $y$ the outer curve has equation $x=\sqrt{y}$.

Similarly, in terms of $y$, the inner curve has equation $x=y^2$.

So the volume is $$\int_{y=0}^1 \pi\left[(\sqrt{y})^2 -(y^2)^2\right]\,dy.$$

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