New Notes Blog: Find the centroid of the region bounded by the graphs of `y=b/asqrt(a^2-x^2)` and `y=0`

Given curves are ,

$\displaystyle{y}={\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}},{y}={0}$

let $\displaystyle{f{{\left({x}\right)}}}={\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}$

and $\displaystyle{g{{\left({x}\right)}}}={0}$

In order to find the Centroid of the region bounded by the curves ,

first we have to find the area bounded by the curves ,so ,

now in order to find the area , we have to find the intersecting points of the curves. This can be obtained by equating f(x) and g(x) .

=> $\displaystyle{f{{\left({x}\right)}}}={g{{\left({x}\right)}}}$

=> $\displaystyle{\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}={0}$

=> $\displaystyle\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}={0}$

=> $\displaystyle{{x}}^{{2}}={{a}}^{{2}}$

=> $\displaystyle{x}=\pm{a}$ ---------(1)

so the curves $\displaystyle{f{{\left({x}\right)}}}\ge{g{{\left({x}\right)}}}{o}{n}{\left[-{a},{a}\right]}$

so the area $\displaystyle={\int_{{a}}^{{b}}}{\left[{f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right]}{\left.{d}{x}\right.}$ where the lower bound is -a, and the upper bound is a

$\displaystyle={\int_{{-{{a}}}}^{{a}}}{\left[{\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}-{0}\right]}{\left.{d}{x}\right.}$

$\displaystyle={\int_{{-{{a}}}}^{{a}}}{\left[{\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right]}{\left.{d}{x}\right.}$

The function which is being integrated is an even function so,

= $\displaystyle{2}{\int_{{0}}^{{a}}}{\left[{\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{2}{\left(\frac{{b}}{{a}}\right)}{\int_{{0}}^{{a}}}{\left[\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right]}{\left.{d}{x}\right.}$

let $\displaystyle{x}={a}{\sin{{\left(\theta\right)}}}$ ------(2)

so , $\displaystyle{\left.{d}{x}\right.}={a}{\cos{{\left(\theta\right)}}}{d}{\left(\theta\right)}$

but from (1) and (2) we get

$\displaystyle{x}=\pm{a},{x}={a}{\sin{{\left(\theta\right)}}}$ is

$\displaystyle{\sin{{\left(\theta\right)}}}=\pm{1}$

=> $\displaystyle\theta={{\sin}}^{{-{1}}}{\left(\pm{1}\right)}$

so $\displaystyle\theta=\pm{\left(\frac{\pi}{{2}}\right)}$

so ,now with the new integrals we get

area $\displaystyle={2}{\left(\frac{{b}}{{a}}\right)}{\int_{{0}}^{{a}}}{\left[\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{2}{\left(\frac{{b}}{{a}}\right)}{\int_{{{0}}}^{{\frac{\pi}{{2}}}}}\sqrt{{{{a}}^{{2}}-{{a}}^{{2}}{{\sin}}^{{2}}{\left(\theta\right)}}}{a}{\cos{{\left(\theta\right)}}}{d}{\left(\theta\right)}$

= $\displaystyle{2}{\left(\frac{{b}}{{a}}\right)}{\int_{{{0}}}^{{\frac{\pi}{{2}}}}}\sqrt{{{{a}}^{{2}}{\left({1}-{{\sin}}^{{2}}{\left(\theta\right)}\right)}}}{a}{\cos{{\left(\theta\right)}}}{d}{\left(\theta\right)}$

= $\displaystyle{2}{\left(\frac{{b}}{{a}}\right)}{\int_{{{0}}}^{{\frac{\pi}{{2}}}}}\sqrt{{{{a}}^{{2}}{\left({{\cos}}^{{2}}{\left(\theta\right)}\right)}}}{a}{\cos{{\left(\theta\right)}}}{d}{\left(\theta\right)}$

= $\displaystyle{2}{\left(\frac{{b}}{{a}}\right)}{\int_{{{0}}}^{{\frac{\pi}{{2}}}}}{\left({a}{\cos{{\left(\theta\right)}}}\right)}{a}{\cos{{\left(\theta\right)}}}{d}{\left(\theta\right)}$

= $\displaystyle{2}{\left(\frac{{b}}{{a}}\right)}{\int_{{{0}}}^{{\frac{\pi}{{2}}}}}{{\left({a}{\cos{{\left(\theta\right)}}}\right)}}^{{2}}{d}{\left(\theta\right)}$

= $\displaystyle{2}{\left(\frac{{b}}{{a}}\right)}{\left({{a}}^{{2}}\right)}{\int_{{{0}}}^{{\frac{\pi}{{2}}}}}{\left({{\cos}}^{{2}}{\left(\theta\right)}\right)}{d}{\left(\theta\right)}$

we can right the above integral as

= $\displaystyle{\left({2}\right)}{\left(\frac{{b}}{{a}}\right)}{\left({{a}}^{{2}}\right)}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({{\cos}}^{{2}}{\left(\theta\right)}\right)}{d}{\left(\theta\right)}$

as we know that $\displaystyle\int{{\cos}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}={\left(\frac{{1}}{{2}}\right)}{\left({x}+{\left(\frac{{1}}{{2}}\right)}{\sin{{\left({2}{x}\right)}}}\right)}$

now ,

area = $\displaystyle{\left({2}\right)}{\left(\frac{{b}}{{a}}\right)}{\left({{a}}^{{2}}\right)}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({{\cos}}^{{2}}{\left(\theta\right)}\right)}{d}{\left(\theta\right)}$

= $\displaystyle{\left({2}\right)}{\left(\frac{{b}}{{a}}\right)}{\left({{a}}^{{2}}\right)}{{\left[{\left(\frac{{1}}{{2}}\right)}{\left({x}+{\left(\frac{{1}}{{2}}\right)}{\sin{{\left({2}{x}\right)}}}\right)}\right]}_{{0}}^{{\frac{\pi}{{2}}}}}$

= $\displaystyle{2}{\left(\frac{{b}}{{a}}\right)}{\left({{a}}^{{2}}\right)}{\left[{\left({\left(\frac{{1}}{{2}}\right)}{\left({\left(\frac{\pi}{{2}}\right)}+{\left(\frac{{1}}{{2}}\right)}{\sin{{\left({2}{\left(\frac{\pi}{{2}}\right)}\right)}}}\right)}\right)}-{\left(\frac{{1}}{{2}}\right)}{\left({\left({0}\right)}+{\left(\frac{{1}}{{2}}\right)}{\sin{{\left({2}{\left({0}\right)}\right)}}}\right)}\right]}$

= $\displaystyle{2}{\left(\frac{{b}}{{a}}\right)}{{a}}^{{2}}{\left[\frac{\pi}{{4}}\right]}$

= $\displaystyle{\left(\frac{{\pi{{a}}^{{2}}}}{{2}}\right)}\cdot{\left(\frac{{b}}{{a}}\right)}$

= $\displaystyle\frac{{\pi\cdot{a}\cdot{b}}}{{{2}}}$

Now the centroid of the region bounded the curves is given as ,

let $\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}$ be the co- ordinates of the centroid so ,

$\displaystyle{x}_{{1}}$ is given as

$\displaystyle{x}_{{1}}={\left(\frac{{1}}{{{a}{r}{e}{a}}}\right)}{\int_{{a}}^{{b}}}{x}{\left[{f{{\left({x}\right)}}}-{g{{\left({x}\right)}}}\right]}{\left.{d}{x}\right.}$

where the limits $\displaystyle{a}=-{a},{b}={a}$

so,

$\displaystyle{x}_{{1}}={\left(\frac{{1}}{{{a}{r}{e}{a}}}\right)}{\int_{{-{{a}}}}^{{a}}}{x}{\left[{\left({\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}-{\left({0}\right)}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{1}}{{\frac{{\pi\cdot{a}\cdot{b}}}{{2}}}}\right)}{\int_{{-{{a}}}}^{{a}}}{x}{\left[{\left({\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{\int_{{-{{a}}}}^{{a}}}{x}{\left[{\left({\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}\right]}{\left.{d}{x}\right.}$

let $\displaystyle{u}={{a}}^{{2}}-{{x}}^{{2}}\Rightarrow{d}{u}=-{2}{x}{\left.{d}{x}\right.}$

$\displaystyle{\left(-\frac{{1}}{{2}}\right)}{d}{u}={x}{\left.{d}{x}\right.}$

The bounds are then $\displaystyle{u}={{a}}^{{2}}-{\left({{a}}^{{2}}\right)}={0}$ and $\displaystyle{u}={{a}}^{{2}}-{{\left(-{a}\right)}}^{{2}}={0}$

so,

$\displaystyle={\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{\int_{{0}}^{{0}}}{\left[{\left({\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}\right]}{\left.{d}{x}\right.}$

$\displaystyle={0}$ since for $\displaystyle{\int_{{a}}^{{b}}}{g{{\left({u}\right)}}}{d}{u}={G}{\left({a}\right)}-{G}{\left({b}\right)}$ it follows that $\displaystyle{\int_{{0}}^{{0}}}{g{{\left({u}\right)}}}{d}{u}={G}{\left({0}\right)}-{G}{\left({0}\right)}={0}$

so, $\displaystyle{x}_{{1}}={0}$

and now let us $\displaystyle{y}_{{1}}$ and so ,

$\displaystyle{y}_{{1}}$ is given as

$\displaystyle{y}_{{1}}={\left(\frac{{1}}{{{a}{r}{e}{a}}}\right)}{\int_{{a}}^{{b}}}{\left(\frac{{1}}{{2}}\right)}{\left[{{f}}^{{2}}{\left({x}\right)}-{{g}}^{{2}}{\left({x}\right)}\right]}{\left.{d}{x}\right.}$

where the $\displaystyle{a}=-{a},{b}={a}$

so ,

= $\displaystyle{\left(\frac{{1}}{{\frac{{\pi\cdot{a}\cdot{b}}}{{{2}}}}}\right)}{\int_{{-{{a}}}}^{{a}}}{\left(\frac{{1}}{{2}}\right)}{\left[{{\left({\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}}^{{2}}-{{\left[{0}\right]}}^{{2}}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{\int_{{-{{a}}}}^{{a}}}{\left(\frac{{1}}{{2}}\right)}{\left[{{\left({\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}}^{{2}}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{\left(\frac{{1}}{{2}}\right)}{\int_{{-{{a}}}}^{{a}}}{\left[{{\left({\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}}^{{2}}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{1}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{\int_{{-{{a}}}}^{{a}}}{\left[{{\left({\left(\frac{{b}}{{a}}\right)}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}}^{{2}}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{1}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{{\left(\frac{{b}}{{a}}\right)}}^{{2}}{\int_{{-{{a}}}}^{{a}}}{\left[{{\left(\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}}^{{2}}\right]}{\left.{d}{x}\right.}$

since the function which is being integrated is even function so ,we can write the above equation as

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{{\left(\frac{{b}}{{a}}\right)}}^{{2}}{\int_{{0}}^{{a}}}{\left[{{\left(\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}\right)}}^{{2}}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{{\left(\frac{{b}}{{a}}\right)}}^{{2}}{\int_{{0}}^{{a}}}{\left[{\left({{a}}^{{2}}-{{x}}^{{2}}\right)}\right]}{\left.{d}{x}\right.}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{{\left(\frac{{b}}{{a}}\right)}}^{{2}}{{\left[{\left({\left({{a}}^{{2}}\right)}{x}-\frac{{{{x}}^{{3}}}}{{3}}\right)}\right]}_{{0}}^{{a}}}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{{\left(\frac{{b}}{{a}}\right)}}^{{2}}{\left[{\left[{\left({\left({{a}}^{{2}}\right)}{a}-\frac{{{{a}}^{{3}}}}{{3}}\right)}\right]}-{\left[{\left({\left({{0}}^{{2}}\right)}{a}-\frac{{{{0}}^{{3}}}}{{3}}\right)}\right]}\right]}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{{\left(\frac{{b}}{{a}}\right)}}^{{2}}{\left[{\left({\left({{a}}^{{3}}\right)}-\frac{{{{a}}^{{3}}}}{{3}}\right)}\right]}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{a}\cdot{b}\right)}}}\right)}{{\left(\frac{{b}}{{a}}\right)}}^{{2}}{\left[\frac{{{2}\cdot{\left({{a}}^{{3}}\right)}}}{{3}}\right)}{\]}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\cdot{b}\right)}}}\right)}{{\left({b}\right)}}^{{2}}{\left[\frac{{{2}}}{{3}}\right]}$

= $\displaystyle{\left(\frac{{2}}{{{\left(\pi\right)}}}\right)}{\left({b}\right)}{\left[\frac{{{2}}}{{3}}\right]}$

= $\displaystyle{\left(\frac{{{4}{b}}}{{{3}\pi}}\right)}$

so the centroid of the area bounded by the curves is

= $\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}={\left({0},\frac{{{4}{b}}}{{{3}\pi}}\right)}$

New Notes Blog

Wednesday, 30 April 2014

Find the centroid of the region bounded by the graphs of $\displaystyle{y}=\frac{{b}}{{a}}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}$ and $\displaystyle{y}={0}$

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Wednesday, 30 April 2014

Find the centroid of the region bounded by the graphs of and

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In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

Find the centroid of the region bounded by the graphs of $\displaystyle{y}=\frac{{b}}{{a}}\sqrt{{{{a}}^{{2}}-{{x}}^{{2}}}}$ and $\displaystyle{y}={0}$

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?