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?

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- Feb 5th 2008, 02:15 PMsimsima_1Centroids
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? - Feb 5th 2008, 03:21 PMwingless
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, $\displaystyle M_x $ and $\displaystyle M_y$, and the area $\displaystyle A$.

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

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

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

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

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

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Let's apply it to your question.

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

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

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

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

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

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

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

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

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

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

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

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

Now remember,

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

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

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

Our centroid is $\displaystyle \text{O }(3.70\text{ , }57.05)$ :) - Feb 5th 2008, 03:59 PMsimsima_1
thank you...i was looking at the question all wrong...i didn't realize that it was bounded by the y-axis.