Answer:
The answer is [tex]\sqrt{\frac{6}{5}}[/tex]
Step-by-step explanation:
To calculate the volumen of the solid we solve the next double integral:
[tex]\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy[/tex]
Solving:
[tex]\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy[/tex]
[tex][6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.[/tex]
Replacing the limits:
[tex]6*\frac{1}{3} =2[/tex]
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 ≤ [tex]\frac{1}{m}[/tex]
Solving the double integral with these new limits we have:
[tex]\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy[/tex]
This part is a little bit tricky so let's solve the integral first for dy:
[tex]\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx[/tex]
Replacing the limits:
[tex]\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx[/tex]
Solving now for dx:
[tex][{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.[/tex]
Replacing the limits:
[tex]\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}[/tex]
As I mentioned before, this volume is equal to 1, hence:
[tex]\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }[/tex]
