Question

Set up and evaluate a double integral to find the volume of the solid bounded by the graphs of the equations. z = xy, z = 0, y = x, x =1, first octant.

Answer 1

1/8

Explanation:

Given:

z = xy

z = 0

y = x

x =1

To find:

volume of the solid bounded by the graphs of the equations

Solution:

Compute integral of volume in the first octant:

`Volume = V = \int\limits^1_0\int\limits^x_0 {z} \, dydx`

`\int\limits^1_0\int\limits^x_0 {z} \, dydx = \int\limits^1_0\int\limits^x_0 {xy} \, dydx`

`= \int\limits^1_0x\int\limits^x_0 {y} \, dydx`

=
`\int\limits^1_0` x y²/2 |ˣ₀ dx

= 1/2
`\int\limits^1_0` x y² |ˣ₀ dx

= 1/2
`\int\limits^1_0` x (x²-0²) dx

= 1/2
`\int\limits^1_0` x³dx

=
`(1)/(2) (x^(3+1) )/(3+1)` |¹₀

= (1/2) (x⁴/4) |¹₀

= 1/8 x⁴ |¹₀

= 1/8 (1⁴ - 0⁴)

= 1/8 (1)

V = 1/8