Hard question

**arze** · February 9th 2010, 10:33 PM

A region of the xy plane is defined by the inequalities
$0\leq y\leq \sin x$ , $0\leq x\leq \pi$

Find:
(a) the area of the region
(b) The first moment of area about the x-axis,
(c) the coordinates of the centroid of this area.

Find Also:
(d)The volume obtained when this area is rotated completely about the x-axis,
(e) The first moment of volume about the y-axis,
(f) The centroid of this volume.

My attempt:
I found (a) correct, its value is 2.
First moment of area about Ox is $Ay\simeq \sum^{\pi}_{x=0} y^2\delta x$
$2y=\int^{\pi}_0(\frac{1}{2}-\frac{1}{2}\cos 2x)dx$
$2y=[\frac{x}{2}-\frac{1}{4}\sin 2x]^{\pi}_0$
$2y=\frac{\pi}{2}$
$y=\frac{\pi}{4}$
This gives me the correct answer, but when i proceed to (c) i have problems. The x-coordinate of the centroid would be $\frac{\pi}{2}$ and the y-coordinate would be $\frac{\pi}{4}$ but the answer is $\frac{\pi}{8}$

I also found (d) correctly it is $\frac{\pi^2}{2}$
Now (e) we have First moment of $\delta V$ about Oy is $\simeq(\pi y^2 \delta x)x$
$Vx\simeq\sum^{\pi}_{x=0}(\pi y^2 x \delta x)$
$\frac{\pi^2}{2}x=\int^{\pi}_0 \pi \sin^3x dx$
$\frac{\pi^2}{2}x=\pi \int^{\pi}_0\sin x(1-\cos^2x)dx$
$\frac{\pi^2}{2}x=\pi [-\cos x+\frac{1}{3}\cos^3x]^{\pi}_0$
$\frac{\pi^2}{2}x=\pi(\frac{4}{3})$
$x=\frac{8}{3\pi}$
answer is $\frac{\pi^3}{4}$
I can do the last one.

Thanks!

**jiboom** · February 10th 2010, 05:11 AM

deleted

**Calculus26** · February 10th 2010, 05:34 AM

For c)

obtains 2y = pi/2 not pi^2/2 as you have x not x^2

so Moment is pi/4

and the the y coordinate is pi/8

**jiboom** · February 10th 2010, 05:52 AM

the first moment $M_y$ is
$ M_y= \int^{\pi}_0(\frac{y^2}{2}) dx$

$ 2M_y=\int^{\pi}_0(\frac{1}{2}-\frac{1}{2}\cos 2x)dx$

$ 2M_y=\frac{\pi}{2} $

so
first moment $M_y= \frac{\pi}{4}$

Y co-ord of centroid is $\frac{M_y}{A}=\frac{\frac{\pi}{4}}{2}=\frac{\pi}{8 }$

for volume:

first moment of volume

$ M_y=\int^{\pi}_0 \pi y^2 x dx$

$ M_y=\int^{\pi}_0 \pi x\sin x^2 dx$

which is

$ \pi( [\frac{x^2}{4}-\frac{x\sin(2x)}{4}-\frac{\cos(2x)}{8}]^{\pi}_0 ) $

$ =\pi([\frac{\pi^2}{4}-0-\frac{1}{8}] -$ $[-\frac{1}{8}]) $

$ =\frac{\pi^3}{4} $

**arze** · February 10th 2010, 10:05 PM

I've got another similar question and this time the borders of the area are y=x^2 and y=4. The parts are about the same.
first i was asked to get the area which i found correctly.
then i am asked to find the first moment of area about the x-axis. so i did this:
$M_x=\int^2_{-2}\frac{(4-x^2)^2}{2}dx=\frac{1}{2}[16x-\frac{8x}{3}+\frac{x^5}{5}]^2_{-2}=\frac{256}{15}$
the answer is $\frac{128}{5}$ as far as i can see my logic is correct and so is the working.

**jiboom** · February 13th 2010, 12:46 PM

Originally Posted by arze

I've got another similar question and this time the borders of the area are y=x^2 and y=4. The parts are about the same.
first i was asked to get the area which i found correctly.
then i am asked to find the first moment of area about the x-axis. so i did this:
$M_x=\int^2_{-2}\frac{(4-x^2)^2}{2}dx=\frac{1}{2}[16x-\frac{8x}{3}+\frac{x^5}{5}]^2_{-2}=\frac{256}{15}$
the answer is $\frac{128}{5}$ as far as i can see my logic is correct and so is the working.

i think you better of integrating between the cure and y -axis here as you have stuff between the curve and x-axis you dont want.
so we want (as we both sides of the y-axis i think the 2 outside integral is ok,it gives the right answer so its promising.

$ 2\int_0^4y(f(y)) $

$ = 2\int_0^4 y(y^{1/2}) $

$ =2 [\frac{2x^{5/2}}{5}]_0^4 $

$ =128/5 $

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