User:ABCD/sandbox2

$$x=ky^2$$ $$y=\sqrt{\frac{x}{k}}$$

$$\bar{x}=\frac{\int_a^b x\cdot\left(y_2\left(x\right)-y_1\left(x\right)\right)dx}{\int^b_a \left(y_2\left(x\right)-y_1\left(x\right)\right)dx}$$ $$\bar{y}=\frac{\int_c^d y\cdot\left(x_2\left(y\right)-x_1\left(y\right)\right)dy}{\int^d_c \left(x_2\left(y\right)-x_1\left(y\right)\right)dy}$$

$$\bar{x}=\frac{\int_0^b x\cdot\left(a-\sqrt{\frac{x}{k}}\right)dx}{\int_0^b\left(a-\sqrt{\frac{x}{k}}\right)dx}$$ $$\bar{y}=\frac{\int_0^a y\cdot ky^2dy}{\int_0^a ky^2dy}$$

$$\bar{x}=\frac{\int_0^b\left(ax-k^{-1/2}x^{3/2}\right)dx}{\int_0^b\left(a-\sqrt{\frac{x}{k}}\right)dx}$$ $$\bar{y}=\frac{\int_0^a ky^3dy}{\int_0^a ky^2dy}$$

$$\bar{x}=\frac{\frac{ab^2}{2}-\frac{5b^{5/2}}{2\sqrt{k}}}{ab-\frac{3b^{3/2}}{2\sqrt{k}}}$$ $$\bar{y}=\frac{\frac{ka^4}{4}}{\frac{ka^3}{3}}$$

$$\bar{x}=\frac{ab\sqrt{k}-5b^{3/2}}{2a\sqrt{k}-3\sqrt{b}}$$ $$\bar{y}=\frac{3a}{4}$$