Area & Centroid of Region Bounded by arcsinx & x=1/2

[whatlifeforme]

Homework Statement

consider the region bounded by the graphs of y=arcsinx, y=0, x = 1/2.

a) find the area of the region.
b) find the centroid of the region.

Homework Equations

$\displaystyle\int_0^{1/2} {arcsinx dx}$

u=arcsinx; du = $\frac{1}{1-x^2}dx$dv=dx ; v=x

xarcsinx]$^{1/2}_{0}$ - $\displaystyle\int_0^{1/2} {\frac{x}{\sqrt{1=x^2}} dx}$

= (1/2arcsin(1/2) - 0) + $\sqrt{1-x^2}^{1/2}_{0}$

answer: $\frac{\pi + 6\sqrt{3}-12}{12}$

The Attempt at a Solution

i can't get that answer.

in my work i have:

xarcsix]$^{1/2}_{0}$ = $\frac{\pi}{12}$for the other part i get:

$\sqrt{1-x^2}^{1/2}_{0}$ = $\sqrt{1-\frac{1}{4}} - 1$which i can see how the last term -1 will go to -12/12, and pi/12 goes if front, but how are the getting the middle term, 6$\sqrt{3}$?

 

[SammyS]

Homework Statement

consider the region bounded by the graphs of y=arcsinx, y=0, x = 1/2.

a) find the area of the region.
b) find the centroid of the region.

Homework Equations

$\displaystyle\int_0^{1/2} {arcsinx dx}$

u=arcsinx; du = $\frac{1}{1-x^2}dx$

dv=dx ; v=x

xarcsinx]$^{0}_{1/2}$ - $\displaystyle\int_0^1/2 {\frac{x}{\sqrt{1=x^2}} dx}$

= (1/2arcsin(1/2) - 0) + $\sqrt{1-x^2}^{1/2}_{0}$

answer: $\frac{\pi + 6\sqrt{3}-12}{12}$

The Attempt at a Solution

i can't get that answer.

in my work i have:

x arcsix]$^{1/2}_{0}$ = $\frac{\pi}{12}$

for the other part i get:

$\sqrt{1-x^2}^{1/2}_{0}$ = $\sqrt{1-\frac{1}{4}} - 1$

which i can see how the last term -1 will go to -12/12, and pi/12 goes if front, but how are the getting the middle term, 6$\sqrt{3}$?

It seems to me that this would be much easier to do by looking at x as a function of y.

 

[iRaid]

I think for this problem, it's just easier to do a substitution at the beginning, x=sinθ. Then you just do a integration by parts.

Note this only works because the limits of integration are between 0 and 1/2. Also, arcsin(sinx)=x.

 

[eumyang]

for the other part i get:

$\sqrt{1-x^2}^{1/2}_{0}$ = $\sqrt{1-\frac{1}{4}} - 1$


which i can see how the last term -1 will go to -12/12, and pi/12 goes if front, but how are the getting the middle term, 6$\sqrt{3}$?

The middle "term" you want to get is $\frac{6\sqrt{3}}{12}$, not $6\sqrt{3}$.

From $\sqrt{1-\frac{1}{4}}$, do the subtraction, and split into two square roots, one in the numerator, and one in the denominator. It's straightforward from there.

 

[whatlifeforme]

solved.thanks.