how to do integration volume question? (1 Viewer)

dan964 · Jun 10, 2015

you could either use a 4U method.

but not necessary.
Just split it into 2 volumes. [Sorry very bad syntax]

Find the volume of y=4-x^2 rotated around and

then subtract from that
the volume of y=4-2x rotated (which happens to be a cone of volume 16pi/3 cubic units)

Edit: Far-out cone not pyramid.

rand_althor · Jun 10, 2015

$\begin{align*} y&=4-x^2 \rightarrow x^2=4-y \\ y&=4-2x \rightarrow 2x=4-y \rightarrow x= \frac{4-y}{2} \rightarrow x^2=\frac{(4-y)^2}{4} \end{align*} \\ \begin{align*} \textup{Volume}&=\pi\int_{0}^{4} (4-y)-\frac{(4-y)^2}{4} \mathrm{d}y \\ &=\pi \left[4y-\frac{y^2}{2} + \frac{(4-y)^3}{12} \right]_0^4 \\ &=\pi \left [\left [4(4)-\frac{4^2}{2} + \frac{(4-4)^3}{12} \right ] - \left [4(0)-\frac{0^2}{2}+\frac{(4-0)^3}{12}) \right ] \right ] \\ &=\pi \left [8-\frac{16}{3} \right ] \\ &=\frac{8}{3}\pi \textup{ units}^3 \end{align*}$

Edit: Fixed an error.

how to do integration volume question? (1 Viewer)

malcolm21

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dan964

what

rand_althor

Active Member

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