Question

find the volume bounded by coordinates plane and the plan x/a+y/b+z/c =1​

Answer 1

The volume bounded by coordinates plane and the plan x/a+y/b+z/c =1​ is abc/6 cubic.units.

To find: the volume bounded by coordinates plane and the plan x/a+y/b+z/c =1​

Given,

The equations of the lines situated on the planes y=0 and z=0 are:

y=0 gives,

x/a+z/c = 1 ⇔ z = (1−x/a)c,

And,

z=0 gives,

x/a+y/c=1 ⇔ y=(1−x/a)b.

Area A(x) is given by,

A(x) = 1/2(1−x/a)b(1−x/a)c

=  bc/2(1−x/a)^2.

Hence the volume is given by the integration of the area A(x) from x = 0 to x = a

$V = \int\limits^a_0 {A(x)} \, dx$

$V =\int\limits^a_0 \dfrac {bc}{2} (1-\dfrac {x}{a})^2 dx$

V = bc/2(a − 2/a × a^2/2 + 1/a^2 × a^3/3)

V = abc/6.