volume of revolution

IBP twice would cover it, but it would be a lot nicer if it was about the y-axis.

Reply 5

OP

Original post by joostan

Is that

\arccos(x)

? If so by parts, though it's a little mean.
If that is

\dfrac{1}{\cos(x)}

then you should recognise the form. Does the question definitely say

x

axis?

it's arc cosx and it is definitely x axis

Reply 6

OP

i got x(arccos(x))² - integral of (-2xarccos(x) / (1-x²)^1/2) for the first by parts but then when I ibp that I don't get an integral at the end which matches the original integral

Reply 7

joostan

13

Original post by bl64

i got x(arccos(x))² - integral of (-2xarccos(x) / (1-x²)^1/2) for the first by parts but then when I ibp that I don't get an integral at the end which matches the original integral

Yup I agree with the first result. But I'm puzzled about your issue - you don't need it to match the original, just press on and you get to an integral which you can easily evaluate.

(edited 8 years ago)

Reply 8

EricPiphany

Alternatively integrating cos(x) around the y-axis using the shell method gives the same result with no substitution, IBP once.

Reply 9

OP

Original post by joostan

Yup I agree with the first result. But I'm puzzled about your issue - you don't need it to match the original, just press on and you get to an integral which you can easily evaluate.

I thought that for these integrals to stop u need an integral at the end to match the original one take it over to the other side and divide your result but the amount which makes the original integral one again

Reply 10

TeeEm

19

this integral is very clever and requires no "donkey work" or beyond Straight maths knowledge ...

Reply 11

OP

how though do I integrate -2x/(1-x²)^1/2 is it going to be -2xarccosx because if so I'll get a repeating integral won't I?

Reply 12

joostan

13

Original post by bl64

how though do I integrate -2x/(1-x²)^1/2 is it going to be -2xarccosx because if so I'll get a repeating integral won't I?

Reverse chain rule.

Reply 13

ghostwalker

Original post by bl64

how though do I integrate -2x/(1-x²)^1/2 is it going to be -2xarccosx because if so I'll get a repeating integral won't I?

You need to use IBP a second time on the form you had in post #7.

If you can't see how, it will help to consider the deriviative of

\sqrt{1-x^2}

Reply 14

OP

Original post by ghostwalker

You need to use IBP a second time on the form you had in post #7.

If you can't see how, it will help to consider the deriviative of

\sqrt{1-x^2}

if I use the one from post 7 and then make u = 2arccosx / (1-x²)^1/2 then if I use the product rule I get -2(1-x²)^1/2 - 2xarccos(x) / (1-x²)^3/2 that's just a mess.

Am I missing something really stupid and obvious?

Reply 15

ghostwalker