Question

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that the plane y = mx divides S into two pieces with equal volume

Answer 1

Answer:

The answer is
`\sqrt{(6)/(5)}`

Explanation:

To calculate the volumen of the solid we solve the next double integral:

`\int\limits^1_0\int\limits^1_0 {12xy^(2) } \, dxdy`

Solving:

`\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^(2) } \, dy`

`[6x^(2) ]{{1} \atop {0}} \right. * [(y^(3))/(3)]{{1} \atop {0}} \right.`

Replacing the limits:

`6*(1)/(3) =2`

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤
`(1)/(m)`

Solving the double integral with these new limits we have:

`\int\limits^(1)/(m) _0\int\limits^(1)_(mx) {12xy^(2) } \, dxdy`

This part is a little bit tricky so let's solve the integral first for dy:

`\int\limits^(1)/(m)_0 [{12x (y^(3))/(3)}]{{1} \atop {mx}} \right.\, dx =\int\limits^(1)/(m)_0 [{4x y^(3 )]{{1} \atop {mx}} \right.\, dx`

Replacing the limits:

`\int\limits^(1)/(m)_0 {4x(1-(mx)^(3) )\, dx =\int\limits^(1)/(m)_0 {4x-4x(m^(3) x^(3) )\, dx =\int\limits^(1)/(m)_0 ({4x-4m^(3) x^(4)) \, dx`

Solving now for dx:

`[{(4x^(2))/(2) -(4m^(3) x^(5))/(5) ]{{(1)/(m) } \atop {0}} \right. = [{2x^(2) -(4m^(3) x^(5))/(5) ]{{(1)/(m) } \atop {0}} \right.`

Replacing the limits:

`(2)/(m^(2) )-(4m^(3)(1)/(m^(5)))/(5) =(2)/(m^(2) )-(4(1)/(m^(2)))/(5) \\ (2)/(m^(2) )-(4)/(5m^(2) )=(10m^(2)-4m^(2) )/(5m^(4)) \\ (6m^(2) )/(5m^(4)) =(6)/(5m^(2))`

As I mentioned before, this volume is equal to 1, hence:

`(6)/(5m^(2))=1\\m^(2) =(6)/(5) \\m=\sqrt{(6)/(5) }`