Question

Can someone please verify my answer to this surface area integral question (Calc II)?

Answer 1

try to type in google search-step by step and then what you want to know

Answer 2

@violabra000 , sorry bud, not that simple.

Answer 3

how did you go from\[\sin(x)\sqrt{\sin^2(x)}\]to\[2\sin(x)\]?

Answer 4

Ahh, you're correct. It should be $$sin^{2}(x)$$

Answer 5

I think that's your only mistake, so long as you do it correct from there

Answer 6

Here is the corrected work

Answer 7

yeah it's correct up to there

Answer 8

Awesome! Thank you.

Answer 9

welcome!

Answer 10

I recommend another approach as I believe what you calculated is the line integral (or the length of the curve, not a surface of the rotated curve). But before I get into my suggestion, just wanted to point out that \[1+\sin^2x \neq cos^2 x\] (this would have been true if there was a minus sign.
Now to the problem: You need to find the surface of a "rotational sinusoid." You can think of that surface as an summation of perimeters of circles with the radius corresponding to their position x, which is \[2\pi \sin x\]. The "summation" is of course a definite integral from 0 to pi:\[\int_0^\pi 2\pi \sin x dx\]. And that's the answer! Have a good one.

Answer 11

yup, I totally messed that one up, thanks for saving it @b87lar

Answer 12

Yes, thank you for catching that mistake - 1-sin(x) = cos(x), not 1+sin(x)

Answer 13

So, why is the square root of 1 + (f'(x))^2 not involved here?

I thought the equation for the surface area of a revolved object was $$SA = \int_{a}^{b} 2 \pi \;y \;ds$$ where $$ds = \sqrt{1 + (f'(y))^{2}} dy$$

Answer 14

yeah it's still got that factor

Answer 15

@LolGoCryAboutIt You're right. Since \(\large y=\sin x\), then \(\large y^{\prime}=\cos x\) and hence \(\large \,ds = \sqrt{1+\cos^2x}\,dx\). Therefore, the surface area integral is
\[\large \int_0^ {2\pi}2\pi \sin x\sqrt{1+\cos^2x}\,dx\]

Answer 16

woops; I just noticed the bounds were from 0 to \(\pi\)

Answer 17

So it's really \(\large \displaystyle \int_0^{\color{red}{\pi}}2\pi\sin x\sqrt{1+\cos^2x}\,dx\). Sorry about that! :-/

Answer 18

So can \[SA = \large \int\limits_0^ {\pi}2\pi \sin(x)\sqrt{1+\cos^2x}\,dx\] be considered the most simplified form?

Answer 19

Since there is no trignometric identity that can reduce $$\sqrt{1+cos^{2}(x)}$$

Answer 20

There's no identity that would allow you to simplify \(\large \sqrt{1+\cos^2 x}\). So the above formula for SA would be your answer since you were told to set up an integral and not evaluate. :-)

If they told you that you had to evaluate it, the next step would be to make the trigonometric substitution \(\large\cos x = \tan\theta\), which makes for quite an amusing calculation. :-)

Answer 21

the integral can be done apparently, but it doesn't seem easy given the indefinite answer
http://www.wolframalpha.com/input/?i=integral%202pi*sin(x)sqrt(1%2Bcos%5E2x)dx&t=crmtb01

Answer 22

Thank you guys very much! I'm really grateful for the peer reviews.