Question

what is the centroid of x^2=y, x+y=6, x=0. with solution

Answer 1

Solution:

$x^2=y, x+y=6 \\ \Rightarrow f(x)=x^2, g(x)=6-x$

$a=-3,b=2$

Now, $A=\int_a^b[f(x)-g(x)]dx$

$A=\int_{-3}^2[x^2-6+x]dx \\=[\dfrac {x^3}3-6x+\dfrac{x^2}2]_{-3}^2 \\=[\dfrac {8}3-12+2]-[-9+18+\dfrac{9}2] \\=-\dfrac{125}6 \\=\dfrac{125}6$ [neglecting negative sign as it is area]

Now, $\bar x=\dfrac 1A\int_a^bx[f(x)-g(x)]dx$

$\bar x=\dfrac {6}{125}\int_{-3}^2x[x^2-6+x]dx \\=\dfrac {6}{125}\int_{-3}^2[x^3-6x+x^2]dx \\=\dfrac {6}{125}[\dfrac{x^4}4-3x^2+\dfrac{x^3}3]_{-3}^2 \\=\dfrac{6}{125}\cdot \dfrac{125}{12} \\=\dfrac{1}{2}$

$\bar y=\dfrac 1A\int_a^b(\dfrac{f(x)+g(x)}2)[f(x)-g(x)]dx \\=\dfrac {6}{125}\int_{-3}^2(\dfrac{x^2+6-x}2)[x^2-6+x]dx \\=\dfrac {3}{125}\int_{-3}^2(x^4-x^2+12x-36)dx \\=\dfrac {3}{125}[\dfrac{x^5}{5}-\dfrac{x^3}{3}+6x^2-36x]_{-3}^2 \\=\dfrac{6}{125}\left(-\dfrac{500}{3}\right) \\=-8$

Thus, centroid is $(\dfrac12,-8)$.