Centroids

**simsima_1** · February 5th 2008, 02:15 PM

Find the centroid of the region bounded by
y=7x^2+3x
y=0
x=0
x=5

Is there an equation i'm suppose to follow? How do i set up the problem?

**wingless** · February 5th 2008, 03:21 PM

Have you learnt how to find centroids? If you have learnt it and you know how to calculate moments, go on and read the solution. If you don't know it yet, there's a nice video on how to find centroids here.

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To find the centroid, we first have to find 3 things. 2 moments, $M_x$ and $M_y$ , and the area $A$ .

If the centroid's coordinates are ( $\overline{x}, \overline{y}$ ),

$\overline{x} = \frac{M_y}{A}$ and $\overline{y} = \frac{M_x}{A}$ .

If a region is bounded by two curves $f(x)$ and $g(x)$ , and by two lines $x=a$ , $x=b$ , then,

$M_y = \int^{b}_{a} x (f(x)-g(x)) \text{dx}$

$M_x = \frac{1}{2} \int^{b}_{a} f(x)^2 - g(x)^2 \text{dx}$

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Let's apply it to your question.

$M_y = \int^{b}_{a} x (f(x)-g(x)) \text{dx}$

$M_y = \int^{5}_{0} x (7x^2 + 3x) \text{dx}$

$M_y = \int^{5}_{0} 7x^3 + 3x^2 \text{dx}$

$M_y = \frac{7x^4}{4} + x^3 |^{5}_{0}$

$M_y = \frac{4875}{4}$

$M_x = \frac{1}{2} \int^{b}_{a} f(x)^2 - g(x)^2 \text{dx}$

$M_x = \frac{1}{2} \int^{5}_{0} (7x^2 + 3x)^2 \text{dx}$

$M_x = \frac{1}{2} \int^{5}_{0} 49 x^4 + 42 x^3 +9 x^2 \text{dx}$

$M_x = \frac{75125}{4}$

$A = \int^{5}_{0} 7x^2 + 3x \text{dx}$

$A = \frac{7x^3}{3} + \frac{3x^2}{2} |^{5}_{0}$

$A = \frac{1975}{6}$

Now remember,
$\overline{x} = \frac{M_y}{A}$ and $\overline{y} = \frac{M_x}{A}$ .

$\overline{x} = \frac{M_y}{A} = \frac{\frac{4875}{4}}{\frac{1975}{6}} = \frac{585}{158} \approx 3.70$

$\overline{y} = \frac{M_x}{A} = \frac{\frac{75125}{4}}{\frac{1975}{6}} = \frac{9015}{158} \approx 57.05$

Our centroid is $\text{O }(3.70\text{ , }57.05)$

**simsima_1** · February 5th 2008, 03:59 PM

thank you...i was looking at the question all wrong...i didn't realize that it was bounded by the y-axis.

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