Answer to Question #229265 in Calculus for Alana

Question

Find Mx and My for the first quadrant area bounded by x^2=y^2=a^2 and the coordinate axes

Answer 1

The set for which you need to find $M_x$ , $M_y$ is $\{(x;y)| 0\leq x\leq a, 0\leq y\leq a\}$ .

$M_x=p\int\limits_a^b\frac{1}{2}[f^2(x)-g^2(x)]dx,$$M_y=p\int\limits_a^bx[f(x)-g(x)]dx$ . In this case $f(x)=a$ and $g(x)=0$ . So $M_x=p\int\limits_0^a \frac{1}{2}a^2dx=\frac{a^2p}{2}(a-0)=\frac{1}{2}pa^3$ and $M_y=p\int\limits_0^a xadx=\frac{1}{2}pa^3$