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C4 integration

Q. Curve C has parametric equations x=sint and y=sin2t, between 0 and pi/2

a. find area between C and x-axis
b. Volume when region is revolved through 2pi radians about the x axis.

thanks for your help.

Reply 1

insparato

Q. Curve C has parametric equations x=sint and y=sin2t, between 0 and pi/2

a. find area between C and x-axis
b. Volume when region is revolved through 2pi radians about the x axis.

Unparseable latex formula:

\displaystyle Area = \int_0^{\frac{\pi}{2}} y \frac{dx}{dt} \hspace5 dt

\frac{dx}{dt} = cost

Unparseable latex formula:

\displaystyle Area = \int_0^{\frac{\pi}{2}} sin2t cost \hspace5 dt

Unparseable latex formula:

= \displaystyle 2 \int cos^2t sint \hspace5 dt

This is a direct recognition integral

= [-\frac{2}{3}cos^3t ]_0^{\frac{\pi}{2}} = 0 - [- \frac{2}{3}] = \frac{2}{3}

x = sint

y = 2sintcost

y = 2x\sqrt{1-x^2}

Unparseable latex formula:

Volume = \displaystyle \pi \int y^2 \hspace5 dx

Unparseable latex formula:

Volume = \displaystyle \pi \int_0^1 4x^2(1-x^2) \hspace5 dx

Unparseable latex formula:

Volume = \displaystyle \pi \int_0^1 4x^2 - 4x^4 \hspace5 dx

Volume = \pi [\frac{4}{3}x^3 - \frac{4}{5}x^5]_0^1 = \pi [\frac{4}{3}-\frac{4}{5}] = \frac{8\pi}{15}

Spoiler

Reply 2

generalebriety

Volume =

\int \pi y^2 \frac{dx}{dt} dt

Reply 3

insparato

generalebriety

Volume =

\int \pi y^2 \frac{dx}{dt} dt

Ah! how stupid of me.

cheers for your help

im really confused. can someone help-

im bumping this thread as im stuck on part a- although the entire answer is written by insparato above i still dont understand how he gets from

int 2cos^2sint to -2/3cos^3t,

obviously hes using the general rule he's written in the spoiler but i dont remember seeing anything about that or direct recognition integrals in my notes-
am i meant to know this for c4?

Reply 6

insparato

I don't think its necessary, you could do a substitution. Direct recognition is really a show of how much experience you've had with integrals. You pick up on certain integrals that crop up from time to time.

\int cos^nx sinx = -\frac{1}{n+1}cos^{n+1}x + C

and

\int sin^nx cosx = \frac{1}{n+1}sin^{n+1}x + C

are two useful ones.

Reply 7

domestic_goddess

ah so its just something you pick up, well those are good to know thanks,
would i have otherwise just done the substitution with u=cosx

Reply 8

generalebriety

insparato

Direct recognition is really a show of how much experience you've had with integrals.

I don't think this is really true, and I say this because I believe it's rather discouraging for people to think "if I can't see the answer without putting pen to paper I'm not good enough". It's simply a measure of how much experience you've had with integrals as long as you've been taught to integrate by recognition. I never was taught to; beyond very simple functions (polynomials, exp, trig, f'/f...), I was taught how to find a good substitution. And it's finding the substitution that became automatic for me, rather than just seeing the answer, simply because that's how I was taught. It's only thanks to TSR (and in particular, you!) that I learnt to do it by recognition - most people I know simply don't know how to do it, not because they're bad at integration, but because they've been taught a much more conservative method.

Reply 9

generalebriety

domestic_goddess

ah so its just something you pick up, well those are good to know thanks,
would i have otherwise just done the substitution with u=cosx

As above. And yes, u = cos x works fine.

Reply 10

domestic_goddess

excellent thanks very much guys :smile:

its good to know that theres still hope for me yet even if i only know substitution :p:

Reply 11

nota bene

For me, before the book started with substitutions it had a couple of exercises on recognition/inspection is simple stuff like x/(x^2+1) - then it went on to teach substitution. But fairly quickly after I understood how to make a substitution I just kind of did it mentally instead, to save time - and after having done this for a while you get good at spotting the integral directly. It's got nothing to do with learning standard integrals - just seeing that something is the derivative of something familiar (at least to me).

I have plenty of friends who need to make a substitution every time, despite being good at integration - but also have friends who do it in the same way as me.