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For such problems of finding volumes of solids enclosed by given graphs, we need to visualize the picture of the solid in our mind.

Here the solid is 1st octant.  It has the base on z = 0  ie., x-y plane.  The base is a triangle enclosed by  y = x, x = 1 and y = 0.

The solid is enclosed with in the planes:  z = 0 at the bottom, x = 1 on the right side, y = 0 on the front side, y = x on the back side. The height of the solid = z = x * y.

Limits for x and y for integration are: 0 <= x <= 1    and  0<= y <= 1    and  0<= z <= 1

In a z plane,
     x = r CosΦ    and  y = r Sin Φ,    Ф is the angle of position vector wrt x axis.
     An area element = dA = dx dy  or in polar coordinates (dr) *(r dΦ)

We find the volume using the formula  dV = z * dA  = x * y * r dr dФ

dV = 1/2 * r³ Sin 2Ф dr dФ

The limits for integration are now:
    r = 1 at  Ф = 0°      and    r = √(1²+1²) = √2    at  Ф = π/4
    Ф = 0°   to  π/4

Volume is now found by double integral:
= \frac{1}{2} \int\limits^{\sqrt2}_{r=1} {} \, {} \int\limits^{\pi/4}_{0} {r^3\ Sin2\phi } \, dr\ d\phi \\\\=\frac{1}{2} \int\limits^{\sqrt2}_{r=1} {r^3} \, dr \int\limits^{\pi/4}_0 {Sin2\phi} \, d\phi \\\\=\frac{1}{8}* [ r^4 ]_1^{\sqrt2} * [ cos2\phi ]_0^{\pi/4} \\\\=\frac{3}{8}

So the volume is  3/8  units.

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