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

Answer:

The volume  $V = \frac{75}{2}$ units

Step-by-step 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  $\{ (x, y) | -1 \le x \le 0 , 1 \le y \le 4 \}$

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    

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

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

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

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

        $V = \int\limits^{4}_{1} 5y \, dy$

        $V = 5 [ \frac{y^2}{2} ]|\left \ 4 } \atop { 1 }} \right. \, dy$

       $V = 5( [ \frac{4^2}{2} ] - [ \frac{1^2}{2} ] )$

        $V = \frac{75}{2}$