# Thread: Surface Area of a Revolution

1. ## Surface Area of a Revolution

Hello all,

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

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

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

2. ## Re: Surface Area of a Revolution

Originally Posted by Ivanator27
Hello all,

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

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

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

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

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

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

3. ## Re: Surface Area of a Revolution

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?

4. ## Re: Surface Area of a Revolution

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.

5. ## Re: Surface Area of a Revolution

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.

6. ## Re: Surface Area of a Revolution

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.

7. ## Re: Surface Area of a Revolution

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

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?

8. ## Re: Surface Area of a Revolution

Originally Posted by Ivanator27
$\large s_{x}=\int_{- \pi}^{ \pi}\sqrt{1+sin^{2}(x)+2xsin(x)cos(x)+x^2cos^{2}(x)}dx$

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.

9. ## Re: Surface Area of a Revolution

Oops, is this better?

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

10. ## Re: Surface Area of a Revolution

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