1
$\begingroup$

So here is my problem:

Find an integral for the area of the surface generated by revolving the curve $y=sin(x)$ between $0 \le x \le \pi$, about the x-axis

Just thinking about the problem I feel like I would just do the following:

$$2\int_{0}^{\pi}\sin(x)\mathrm{d}x$$

But I'm not sure if this is the way to go about doing this. Any hints?

$\endgroup$
1
  • $\begingroup$ Write $\sin$ using \sin instead of sin and write $\mathrm dx$ using \mathrm{d}x instead of dx. $\endgroup$
    – user93957
    Dec 7 '13 at 21:38
1
$\begingroup$

The surface area is found by the formula $$A = 2\pi\int\rho\, ds,$$ where $\rho$ is the radius of rotation and $ds$ is the element of arc-length. You have $$A = 2\pi \int_0^\pi \sin(x)\sqrt{1 + \cos^2(x)}\, dx$$ Can you integrate this?

$\endgroup$
1
  • $\begingroup$ If the OP can't,he or she should consider another line of study because that integrand's about as easy as it gets from here.........lol $\endgroup$ Dec 28 '17 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.