Find the centroid $(\bar{x}, \bar{y})$ of the region bounded by: $y = 6x^2+7x$, $y = 0$, $x = 0$, and $x = 7$.
Can some one help with part y.c because webwork say it wrong.
this is my work:
$$A = \int_0^7 y_1 – y_2\,\mathrm{d}x = \int_0^7 6x^2 +7x – 0 \,\mathrm{d}x = \int_0^7 6x^2 + 7x \,\mathrm{d}x$$
Integrating:
$$A = \left[2x^3 + \frac{7x^2}{2}\right]_0^7$$
Evaluating:
$$A = \left[2(7)^3 + \frac{7(7)^2}{2}\right]- \left[2(0)^3 + \frac{7(0)^2}{2}\right]$$
$$A = \frac{1715}{2} \mathrm{\,units}^2$$
The $x$-centroid, $x_c$, is the distance from the $y$-axis to the centroid of the area:
$$x_c =\frac{1}{A} \int_0^7 x\left(y_1 – y_2\right)\,\mathrm{d}x =\frac{1}{A} \int_0^7 x\left(6x^2 +7x – 0\right) \,\mathrm{d}x =\frac{1}{A} \int_0^7 6x^3 + 7x^2 \,\mathrm{d}x$$
Integrating:
$$x_c =\frac{1}{A}\left[\frac{3}{2}x^4 + \frac{7}{3}x^3\right]_0^7$$
Evaluating:
$$x_c =\frac{1}{\frac{1715}{2}}\left[\left[\frac{3}{2}(7)^4 + \frac{7}{3}(7)^3\right]-\left[\frac{3}{2}(0)^4 + \frac{7}{3}(0)^3\right]\right]$$
$$x_c =\frac{77}{15}\approx5.133\mathrm{\,units}$$
The $y$-centroid, $y_c$, is the distance from the $x$-axis to the centroid of the area:
$$y_c = \frac{1}{A} \int_0^7 \left(y_1 – y_2\right)^2\,\mathrm{d}x = \frac{1}{A} \int_0^7 \frac{1}{2}(6x^2 + 7x – 0)^2 \,\mathrm{d}x = \frac{1}{A} \int_0^7 (3x^2 + \frac{7}{2}x)^2 \,\mathrm{d}x$$
Expanding:
$$y_c = \frac{1}{A} \int_0^7 (9x^4 + 21x^3 + \frac{49}{9}x^2) dx$$
Integrating:
$$y_c = \frac{1}{A} \left[\frac{9}{5}x^5 + \frac{21}{4}x^4 + \frac{49}{12}x^3\right]_0^7$$
Evaluating:
$$y_c = \frac{1}{\frac{1715}{2}} \left[\left[\frac{9}{5}(7)^5 + \frac{21}{4}(7)^4 + \frac{49}{12}(7)^3\right]-\left[\frac{9}{5}(0)^5 + \frac{21}{4}(0)^4 + \frac{49}{12}(0)^3\right]\right]$$
$$y_c = \frac{3871}{75} ≈ 51.613 \mathrm{\,units}$$
Best Answer
The region we are talking about is the region under the curve $y = 6x^2 + 7x$ between the points $x = 0$ and $x = 7$. Consider this region to be a laminar sheet. In that case, the centroid is the centre of mass of the lamina.
$$ y_{\text{cm}} = \frac{\int y \ dm}{\int dm}, \ x_{\text{cm}} = \frac{\int x \ dm}{\int dm}. $$
Let $\sigma = \text{mass per unit area}$.
The abscissa can thus be calculated as:
$$ x_{\text{cm}} = \frac{\int_0^7 \sigma xy \ dx}{\int_0^7 \sigma y \ dx} = \frac{\int_0^7 xy \ dx}{\int_0^7 y \ dx}. $$
Now,
$$ \frac{dy}{dx} = 12x + 7 \\ \implies dy = (12x + 7) dx. $$
The ordinate can thus be calculated as:
$$ y_{\text{cm}} = \frac{\int \sigma y(7 - x) \ dy}{\int \sigma (7-x) \ dy} \\ = \frac{\int_0^7 y(7-x)(12x + 7) \ dx}{\int_0^7 (7-x)(12x + 7) \ dx}. $$