Question

Find the volume of the solid lying under the plane z 2x 5y 1 and above the rectangle x, y 1 x 0 1, .

Answer 1

The volume
`V = (75)/(2)` units

Explanation:

From the question we are told that

The equation of the plane is
`z = 2x + 5y + 1`

The range of the rectangle on the x and y axis is
`\ -1 \le x \le 0 , 1 \le y \le 4 \`

Generally the volume is mathematically represented as

`V = \int\limits^(4)_(1) \int\limits^0_(-1) {z} \, da`

Here stands for the area of the solid

So

`V = \int\limits^(4)_(1) \int\limits^0_(-1) {(2x + 5y + 1 )} \, dxdy`

`V = \int\limits^(4)_(1) [{((2x^2)/(2) + 5xy + x )} ] | \left \ 0} \atop {-1}} \right. \, dy`

`V = \int\limits^(4)_(1) [{(x^2 + 5xy + x )} ] | \left \ 0} \atop {-1}} \right. \, dy`

`V = \int\limits^(4)_(1) [{((0)^2 + 5(0)y + (0) )} ] -[{((-1)^2 + 5(-1)y + (-1) )} ] \, dy`

`V = \int\limits^(4)_(1) 5y \, dy`

`V = 5 [ (y^2)/(2) ]|\left \ 4 } \atop { 1 }} \right. \, dy`

`V = 5( [ (4^2)/(2) ] - [ (1^2)/(2) ] )`

`V = (75)/(2)`