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**jstout64** Find te volume of the solid in the first octant of XYZ space bounded below by the coordinate axes and the unit circle, and bounded above by z = 8xy.

So i want the double integral of 8xy dy dx.

The outer integral should be simple, x goes from 0 to 1.

The inner integral though we need to solve for y. y^2 + x^2 = 1.

So y = + or - sqrt(1 - x^2) I only want the positive root.

But since its in the first octant should it be sqrt(1-x^2)/2? because i only want the quarter circle... So my limits for y should be 0 to sqrt(1-x^2)/2.

Do i need to divide by 2?