Surface Area of a Revolution

**Ivanator27** · May 12th 2012, 12:18 PM

Hello all,

I need to find the surface area of the revolution of
$y=xsin(x)$
from -pi to pi about the x-axis. I've done this so far:

$\frac{dy}{dx}=sin(x)+xcos(x)$
$s_{x}=\int_{- \pi}^{ \pi}\sqrt{1+sin^{2}(x)+2xsin(x)cos(x)+x^2cos^{2}(x)}dx$

I've tried a lot to simplify this and I've failed. Any ideas? Have I done something wrong?

**skeeter** · May 12th 2012, 12:45 PM

Originally Posted by Ivanator27

Hello all,

I need to find the surface area of the revolution of
$y=xsin(x)$
from -pi to pi about the x-axis. I've done this so far:

$\frac{dy}{dx}=sin(x)+xcos(x)$
$s_{x}=\int_{- \pi}^{ \pi}\sqrt{1+sin^{2}(x)+2xsin(x)cos(x)+x^2cos^{2}(x)}dx$

I've tried a lot to simplify this and I've failed. Any ideas? Have I done something wrong?

surface area of rotation ...

$S = 2\pi \int_a^b r(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

$S = 2\pi \int_{-\pi}^\pi x\sin{x} \sqrt{1 + \left(\sin{x}+x\cos{x}\right)^2} \, dx$

I don't think this integral can be simplified using elementary methods ... may have to do the calculation w/ a machine

wolfram says ...

definite integral of x*sin(x)*sqrt(1+(sin(x)+x*cos(x))^2) from -pi to pi - Wolfram|Alpha

**Ivanator27** · May 12th 2012, 01:43 PM

Hmmmm... is there not some elementary basis for finding the surface area of a revolution? Like the Riemann Sum for area under a curve. Perhaps I could use that?

**skeeter** · May 12th 2012, 01:48 PM

Is there some reason not to use technology? If you insist, go with a Riemann sum ... however, you'll most probably have to use technology to calculate that, also.

**Ivanator27** · May 14th 2012, 06:32 AM

I would like to know the principles involved with solving this problem, I don't mind using a computer or calculator if I understand what it's doing.

**HallsofIvy** · May 14th 2012, 06:52 AM

skeeter is suggesting a numerical integration. What you mention, the "Riemann sum", where you approximate the "area under the curve" as a rectangle (the function is approximated by piecewise constant), is one possibility. Another is the "trapezoid" method where you approximate the area by trapezoids (the function is approximated by piecewise linear functions). A third is "Simpson's rule" where the function is approximated by piecewise quadratic functions. You should be able to find all of those in any Calculus textbook.

**Ivanator27** · May 14th 2012, 07:08 AM

$s_{x}=\int_{- \pi}^{ \pi}\sqrt{1+sin^{2}(x)+2xsin(x)cos(x)+x^2cos^{2}(x)}dx$

I'm familiar with Riemann sums and trapezium rule, so can I just use a Riemann sum to evaluate the expression being integrated? i.e. can I just do a Riemann sum using $\sqrt{1+sin^{2}(x)+2xsin(x)cos(x)+x^2cos^{2}(x)}$ and the bounds pi and -pi? If i do so, could I post my working here so someone could check it?

**skeeter** · May 14th 2012, 07:15 AM

Originally Posted by Ivanator27

$s_{x}=\int_{- \pi}^{ \pi}\sqrt{1+sin^{2}(x)+2xsin(x)cos(x)+x^2cos^{2}(x)}dx$

I'm familiar with Riemann sums and trapezium rule, so can I just use a Riemann sum to evaluate the expression being integrated? i.e. can I just do a Riemann sum using $\sqrt{1+sin^{2}(x)+2xsin(x)cos(x)+x^2cos^{2}(x)}$ and the bounds pi and -pi? If i do so, could I post my working here so someone could check it?

note that your integral represents the arc length of the curve, not the surface area of revolution as was stated in your original post.

**Ivanator27** · May 14th 2012, 07:22 AM

Oops, is this better?

$s_{x}=\int_{- \pi}^{ \pi}xsin(x)\sqrt{1+sin^{2}(x)+2xsin(x)cos(x)+x^2cos^{2}(x)}dx$

**skeeter** · May 14th 2012, 07:41 AM

Originally Posted by Ivanator27

Oops, is this better?

$s_{x}=\int_{- \pi}^{ \pi}xsin(x)\sqrt{1+sin^{2}(x)+2xsin(x)cos(x)+x^2cos^{2}(x)}dx$

see post #2

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