Hi, Can some one Please help. centre mas question
The 'question' is in blue on my drawing, I have attempted it but I dont believe it is correct ,I made up my own value for variable c , as they didnt give values
(I dont really understand it)
Hi, Can some one Please help. centre mas question
The 'question' is in blue on my drawing, I have attempted it but I dont believe it is correct ,I made up my own value for variable c , as they didnt give values
(I dont really understand it)
$\displaystyle \bar{y} = \frac{\int_{-c}^c x^4 \, dx}{2\int_{-c}^c x^2 \, dx}
$
taking advantage of symmetry ...
$\displaystyle \bar{y} = \frac{\int_0^c x^4 \, dx}{2\int_0^c x^2 \, dx}
$
$\displaystyle \bar{y} = \frac{\frac{c^5}{5}}{\frac{2c^3}{3}} = \frac{3c^2}{10}
$
so, if c = 3 , then the y-coordinate of the region's centroid is $\displaystyle \bar{y}= \frac{27}{10}$ , and the x-coordinate is $\displaystyle \bar{x} = 0$ as evidenced by the function's symmetry to the y-axis.
They are saying that your region is bounded by the parabola $\displaystyle y=x^2$ and the two vertical lines $\displaystyle x = -c$ and $\displaystyle x = c$ where $\displaystyle c$ is just some positive constant (the $\displaystyle c>0$ part).
So when you set up your integral to find $\displaystyle \bar{y}$, we have: $\displaystyle \bar{y} = \frac{\int_{-c}^c x^4 \, dx}{2\int_{-c}^c x^2 \, dx}$
Now, if $\displaystyle c=3$, then we would have: $\displaystyle \bar{y} = \frac{\int_{-3}^3 x^4 \, dx}{2\int_{-3}^3 x^2 \, dx}$
You can solve that directly if you want. But what we noted earlier is that both $\displaystyle y= x^4$ and $\displaystyle y = x^2$ are even functions, and using the fact I gave you in my last post, we can convert each integral so that we have:
$\displaystyle \bar{y} = \frac{\int_{-3}^3 x^4 \, dx}{2\int_{-3}^3 x^2 \, dx} = \frac{{\color{red}2}\int_{{\color{red}0}}^3 x^4 \, dx}{2 \cdot {\color{red}2}\int_{{\color{red}0}}^3 x^2 \, dx} = \frac{\int_{0}^3 x^4 \, dx}{2\int_{0}^3 x^2 \, dx}$
Doesn't really do anything too much but it makes evaluating the definite integral a little easier. We like zeroes .. kind of..
-c did not become zero.
Both functions, $\displaystyle x^4$ and $\displaystyle x^2$ are symmetrical to the y-axis ... as such, I took advantage of the symmetry by writing $\displaystyle \int_{-c}^c x^4 \, dx = 2\int_0^c x^4 \, dx$ and $\displaystyle \int_{-c}^c x^2 \, dx = 2\int_0^c x^2 \, dx$ .