Answer to Question #142086 in Calculus for Promise Omiponle

Question

Find the centroid (x¯,y¯) of the region bounded by:
y=7x^2+6x, y=0, x=0, and x=6.

Answer 1

$Solution$

Given $y=7x^2+6x,y=0,x=0,x=6$

$dA=7x^2+6x\\ \bar x=\frac{\intop_0^6 x dA}{\intop_0^6 dA}=\bar x=\frac{\intop_0^6 x (7x^2+6x)dx}{\intop_0^6 (7x^2+6x)dx}\\ =\frac{\intop_0^6 7x^3dx+\intop_0^6 6xdx}{\intop_0^6 7x^2dx+\intop_0^66xdx}=\frac{[\frac{7x^4}{4}]_0^6+[\frac{6x^2}{2}]_0^6}{[\frac{7x^3}{3}]_0^6+[\frac{6x^2}{2}]_0^6}=\frac{[\frac{7}{4}(6^4-0)]+[\frac{6}{2}(6^4-0)]}{[\frac{7}{3}(6^4-0)]+[\frac{6}{2}(6^4-0)]}\\ \bar x=\frac{2268+108}{504+108}=\frac{2376}{612}=3.88\\$

$\bar y=\frac{\intop_0^6 y dA}{\intop_0^6 dA}=\bar y=\frac{\intop_0^6 (7x^2+6x)^2dx}{\intop_0^6 (7x^2+6x)dx}\\ =\frac{\intop_0^6 49x^4dx+84x^3+36x^2}{\intop_0^6 7x^2dx+\intop_0^66xdx}=\frac{49(\frac{x^5}{5})_0^6+84(\frac{x^4}{4})_0^6+36(\frac{x^3}{3})_0^6}{[\frac{7x^3}{3}]_0^6+[\frac{6x^2}{2}]_0^6}=\frac{49(\frac{6^5}{5})+84(\frac{6^4}{4})+36(\frac{6^3}{3})}{[\frac{7(6)^3}{3}]+[\frac{6(6)^2}{2}]}\\ \bar y=\frac{76204.8+27216+2592}{504+108}=\frac{106012.8}{612}=1732.23\\ \therefore (\bar x, \bar y)=(3.88, 1732.23)$