# Find the volume of the solid in the first octant bounded by the graphs z=1-(y^2), y=2x and x=3.

You need to use triple integral to evaluate the volume of solid such that:

`dV = dxdydz`

Notice that the solid is in first octant, hence, `xgt0, ygt0,zgt0` . Since `zgt0 and ygt0 =gt z = y^2 - 1 gt 0 `

`V = int_0^3 dx int_0^(2x) dy int_0^(y^2-1)dz`

You need to evaluate the inner integral `int_0^(y^2-1)dz`  such that:

`int_0^(y^2-1)dz = z |_0^(y^2-1)`

`int_0^(y^2-1)dz = y^2-1`

You need to evaluate the middle integral such that:

`int_0^(2x) (y^2-1) dy = int_0^(2x) y^2dy - int_0^(2x) dy`

`int_0^(2x) (y^2-1) dy = (y^3/3 - y)|_0^(2x)`

`int_0^(2x) (y^2-1) dy = (8x^3)/3 - 2x`

You need to evaluate the outer integral such that:

`int_0^3 ((8x^3)/3 - 2x)dx =int_0^3 (8x^3)/3 dx - int_0^3 2xdx`

`int_0^3 ((8x^3)/3 - 2x) dx = ((8/3)(x^4/4) - x^2)|_0^3`

`int_0^3 ((8x^3)/3 - 2x) dx = (2x^4/3 - x^2)|_0^3`

`int_0^3 ((8x^3)/3 - 2x) dx = 2*27 - 9 = 45`

Hence, evaluating the volume of solid in the first octant yields V = 45.

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