Hello all,
I need to find the surface area of the revolution of
from -pi to pi about the x-axis. I've done this so far:
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
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.
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 and the bounds pi and -pi? If i do so, could I post my working here so someone could check it?