Question

Given the region bounded by the graphs y=xsin(x), y=0, x=0, x=pi, find a) the volume of the solid generated by revolving the region about the x-axis, b) the volume of the solid generated by revolving the region about the y-axis, c) the centroid of the region (assume p=1) by using disk/washer/shell methods and integration by parts.

Answer 1

which part are you having issues with?

Answer 2

x in the front for both

Answer 3

y = xsin(x) from 0 to pi, revolved around the x axis tells us that the radius of each piece is equal to y

the area of each slice of rotation is: pi r^2\[\int_{0}^{pi}~pi~(xsin(x))^2~dx\]
\[pi \int_{0}^{pi}~x^2~sin^2(x)~dx\]
my first thought is a by parts setup

Answer 4

of using a trig identity on the sin^2

cos(2a) = cos^2(a) - sin^2(a)
= (1-sin^2(a)) - sin^2(a)
= 1 -2sin^2(a)

cos(2a) - 1= -2sin^2(a)
\[\frac12-\frac12cos(2a)=sin^2(a)\]
\[\int \frac{x^2}2-\frac{x^2}2cos(2x)~dx\]
might make for an easier run

Answer 5

we can table this out as:

-cos(2x)/2
x^2 -sin(2x)/4 |
- 2x cos(2x)/8 | multiply these rows and add them together
+ 2 sin(2x)/16 |

\[\int -\frac{x^2}2cos(2x)~dx=-\frac{x^2}{4}sin(2x)-\frac{x}{4}cos(2x)+\frac18sin(2x)\]
if i didnt make an error

Answer 6

so what happens to the X^2

Answer 7

oh okay i got it yup thank u

Answer 8

so what will you do for the y-axis

Answer 9

the y axis would require us to look at the function as x = f(y), which is the inverse, but that may require some extra thought

Answer 10

i believe the shell method would be better for the rotation around the y axis

Answer 11

the area of the walls of a cylinder (a shell) is 2pi r from the base circumference, times the height:

Area = 2pi r h

in this case, r relates to x, and h relates to x sinx\[\int_{0}^{pi}~2pi~x(xsin(x))~dx\]

Answer 12

ugh, i did x and not x^2 lol

Answer 13

sin(x)
x^2 -cos(x)
- 2x -sin(x)
+ 2 cos(x)

at 0 we get -2
at pi we get pi^2 - 2

2pi(pi^2-4)

Answer 14

so what you just did is the the yaxis

Answer 15

and for the centroid

Answer 16

i believe centroid refers to a point that balances the whole area on the tip of a pin ... if that makes any sense.

assuming the density is uniform thruout, you will want to know the total area, which is just an integration of x sinx from 0 to pi, a rather simple by parts

knowing this, you can solve for half of it:
\[Area:\int_{0}^{pi}f(x)~dx=F(pi)-F(0)\]
\[\frac12Area:\int_{0}^{b}f(x)~dx=F(b)-F(0)\]
\[F(b)=\frac12Area-F(0)\]
and solve for b

Answer 17

this would give you x=b as the balance point along that dimension, still pondering the y tho

Answer 18

can you plz finish it

Answer 19

i would perfer to see some of your work on that last explanation i did; the integration itself is pretty simple enough to do.

as for the y axis, im thinking we could need to develop a strategy of subtracting a constant from the function to determine a half area

Answer 20

but i cant see a good way to determine that, we would have to parse it another way, split it into 3 sections such that
\[Area=\int_{0}^{c}xsin(x)~dx+\int_{c}^{d}xsin(x)~dx+\int_{d}^{pi}xsin(x)~dx\]
but i havent got a real good idea for it yet

Answer 21

so why did u do it 3 times

Answer 22

i was just considering something that came to mind, but i cant see a way that it makes it simpler in this instance, so i have to re-examine it

Answer 23

the area is equal to pi. so half of the area is pi/2 but im running into a blank solving for:

sin(b) - b cos(b) = pi/2

Answer 24

i got no good ideas coming to me for the centroid :/ you have any ideas?