i have done the part a, for b , i use the key in the (circled part equation ) in to calculator .. my ans is also different form the ans given. is my concept correct by the way? Attachment 30804Attachment 30805Attachment 30806

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- May 1st 2014, 01:08 AMdelsovolume of revolution
i have done the part a, for b , i use the key in the (circled part equation ) in to calculator .. my ans is also different form the ans given. is my concept correct by the way? Attachment 30804Attachment 30805Attachment 30806

- May 1st 2014, 02:26 AMMINOANMANRe: volume of revolution
I haven't see in your calculations to multiply by π .

I found after integration from 0 to 1/2 V = 5.0477...

check it and do not forget to multiply by π. - May 1st 2014, 06:11 AMdelsoRe: volume of revolution
ok, noted my mistake

- May 1st 2014, 07:13 AMdelsoRe: volume of revolution
i redo the question, now got stuck here... which part is wrong?Attachment 30810

- May 1st 2014, 03:52 PMProve ItRe: volume of revolution
For one thing, a volume by revolution is given by $\displaystyle \begin{align*} V = \pi \int_a^b{\left[ f(x) \right] ^2 \, dx} \end{align*}$. You still don't have $\displaystyle \begin{align*} \pi \end{align*}$...

- May 2nd 2014, 02:09 AMdelsoRe: volume of revolution
using calcultor, before times pi, the ans should be 1.60675, by my ans only 1.5... which part is wrong? this almost drive me crazy!

- May 2nd 2014, 03:25 AMProve ItRe: volume of revolution
$\displaystyle \begin{align*} V &= \pi \int_0^{\frac{1}{2}}{ \left( 1 + \frac{1}{4x^2 + 1} \right) ^2 \, dx } \\ &= \pi \int_0^{\frac{\pi}{4}}{ \left\{ 1 + \frac{1}{4 \left[ \frac{1}{2}\tan{(\theta)} \right] ^2 + 1 } \right\} ^2 \, \frac{1}{2}\sec^2{(\theta)}\,d\theta } \textrm{ after making the substitution } 2x = \tan{(\theta)} \\ &= \frac{\pi}{2}\int_0^{\frac{\pi}{4}}{ \left[ 1 + \frac{1}{\tan^2{(\theta)} + 1 } \right] ^2 \sec^2{(\theta)}\,d\theta } \\ &= \frac{\pi}{2}\int_0^{\frac{\pi}{4}}{ \left[ 1 + \frac{1}{\sec^2{(\theta)} } \right] ^2 \sec^2{(\theta)}\,d\theta} \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{4}}{ \left[ 1 + \cos^2{(\theta)} \right] ^2 \sec^2{(\theta)} \,d\theta } \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{4}}{ \left[ 1 + 2\cos^2{(\theta)} + \cos^4{(\theta)} \right] \sec^2{(\theta)}\,d\theta } \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{4}}{ \sec^2{(\theta)} + 2 + \cos^2{(\theta)} \,d\theta } \\ &= \int_0^{\frac{\pi}{4}}{ \sec^2{(\theta)} + 2 + \frac{1}{2} + \frac{1}{2}\cos{(2\theta)} \,d\theta } \\ &= \frac{\pi}{2} \int_0^{\frac{\pi}{4}}{ \sec^2{(\theta)} + \frac{5}{2} + \frac{1}{2}\cos{(2\theta)} \,d\theta } \\ &= \frac{\pi}{2} \left[ \tan{(\theta)} + \frac{5}{2}\theta + \frac{1}{4}\sin{(2\theta)} \right]_0^{\frac{\pi}{4}} \\ &= \frac{\pi}{2} \left\{ \left[ \tan{ \left( \frac{\pi}{4} \right) } + \frac{5}{2}\left( \frac{\pi}{4} \right) + \frac{1}{4}\sin{ \left( \frac{\pi}{2} \right) } \right] - \left[ \tan{(0)} + \frac{5}{2} \left( 0 \right) + \frac{1}{4}\sin{(0)} \right] \right\} \\ &= \frac{\pi}{2} \left[ \left( 1 + \frac{5\pi}{8} + \frac{1}{4} \right) - \left( 0 + 0 + 0 \right) \right] \\ &= \frac{\pi}{2} \left( \frac{5}{4} + \frac{5\pi}{8} \right) \\ &= \frac{5\pi}{8} \left( 1 + \frac{\pi}{2} \right) \\ &= \frac{5\pi}{16} \left( 2 + \pi \right) \end{align*}$

- May 2nd 2014, 05:32 AMdelsoRe: volume of revolution
thank you very much... i should have done in this way instead of expand the y^2

- May 2nd 2014, 06:35 AMProve ItRe: volume of revolution