Find the volume of the given solid. Bounded by the coordinate planes and the plane 5x + 3y + z = 15. 

Solution:

Given the plane 5x + 3y + z = 15

Divide both side with “15”

x/3 + y/5 + z/15 = 1

x: 0 to 3; y: 0 to 5 - (5x/3); z: 0 to 15 - 3y - 5x

Volume of a solid can be given as a triple integral.

Volume = ∫∫∫ dV

We know that dV= dxdydz

Since it is a product, we can rearrange the terms as dz.dy.dx

V = ∫∫∫ dz dy dx

First, solve the innermost derivative dz

V = ∫∫z|\(_0^{15- 3y - 5x}\) dy dx [since ∫1.dx = x ]

V = ∫∫(15- 3y - 5x) dy dx [by applying the limits]

Now integrate w.r.t ‘y’ keeping ‘x’ constant

V =[∫∫{15y - 3y²/2 - 5xy }]\(_0^{5 - 5x/3}\)dx

V = ∫{15(5 - 5x/3) - 3(5 - 5x/3)²/2 - 5x(5 - 5x/3)}dx

V = ∫{75 - 75x/3 - 3(25 - 50x/3 + 25x²/9) /2 - 25x + 25x²/3} dx

V = ∫{75 - 75x/3 - 75/2 + 50x/2 - 25x/6 - 25x + 25x²/3} dx

V = ∫{75(6) - 75(2)x - 75(3) + 50x(3) - 25x - 25x(6) - 25(2)x²}/6} dx

V = ∫{ -50x² - 150x +150x - 25x - 150x + 75(6 - 3)} /6dx

V = ∫{-50x² - 175x + 75(3)} /6} dx

V= ∫\(_0^3\)-25x²/3 - 175x/6 + 75/2} dx

V= [-25x³/3(3) - 175x²/6(2) + 75x/2]\(_0^3\)

V = -25(3)³/9 - 175(3)²/12 + 75(3)/2

V = -75 - 175(3)/4 + 75(3)/2

V = -75 - 131.25 + 112.5

V = -206.25 + 112.5

V = -93.75 cubic units

Volume of the solid is 93.75 cubic units

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Find the volume of the given solid. Bounded by the coordinate planes and the plane 5x + 3y + z = 15.

Summary:

The volume of the given solid. Bounded by the coordinate planes and the plane 5x + 3y + z = 15 is 93.75cubic units.