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Find volume of solid bounded by graphs of the equations z=xy, z=0, x=y, x=1 in the first octant

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by Naima92

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by Naima92

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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:

*So the volume is 3/8 units.*

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

An area element = dA = dx dy or in polar coordinates

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: