Need Integration help! (1 Viewer)

wootwoot1234 · May 1, 2012

Hey guys,

Could someone please give me some help with the following Integration questions:

Please include working out where possible!

Thank you very much

wootwoot1234 · May 1, 2012

Please, anyone?

SpiralFlex · May 1, 2012

$Clearly if we find the volume bounded by the curve$ \;y=x^2 \; $and the x axis and add onto the volume vounded by the curve$\; (x-2)^2 \;$and the x axis, we would have out total volume.$

$V=\pi \int^{1}_{0} x^4 \;dx+ \pi \int^{2}_{1} (x-2)^4\;dx$

$The volume can be found by subtracting the volume of the curve closest from the axis of rotation from the curve furthest away from the axis of rotation.$

$Note: They intersect at the point (2,4) and (-1,1)$

$V=\pi \int^{2}_{-1}((x+2)^2-x^2)\;dx$

$V=\pi \int^{r}_{-r}(r^2-x^2)\;dx$

$By symmetry,$

$V=2\pi \int^{r}_{0}(r^2-x^2)\;dx$

$=2\pi [r^2x - \frac{x^3}{3}]_{0}^{r}$

$=2\pi[r^3-\frac{1}{3}r^3]-2\pi*[0]$

$\therefore V = \frac{4}{3}\pi r^3 \; \textup{units}^3$

Try integrating now.

Drongoski · May 1, 2012

Q21)

V = V1 + V2

$\int _0 ^1 \pi y^2 dx + \int _1 ^ 2 \pi y^2 dx$

$= \pi [\int _0 ^1 (x^2)^2 dx + \int _1 ^2 ((x-2)^2)^2 dx]$

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or by symmetry 2 x V1.

ok - spiral can complete it.

wootwoot1234 · May 1, 2012

Thanks for your help Spi and Drongoski!

Need Integration help! (1 Viewer)

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