Help with Integration questions (1 Viewer)

calebwhitaker199 · Nov 11, 2015

Hey people, just looking for the process to work these problems out, thank you in advance.

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InteGrand · Nov 11, 2015

calebwhitaker199 said:
Hey people, just looking for the process to work these problems out, thank you in advance.

Sent from my iPad using Tapatalk

$In general, sketch the curves to see what the required region looks like and find the $x$-values of the points of intersections of the curves so you know what to use for the limits of integration. If the function whose curve is further from the $x$-axis (i.e. higher up, if both curves are above the $x$-axis) is $f(x)$ and the function that closer to the $x$-axis (i.e. is lower if both curves are above the $x$-axis) is $g(x)$, then the volume formed by rotating the area between these two curves about the $x$-axis is $V=\pi \int _a ^b \Big{(}\left(f(x)\right)^2 - \left(g(x)\right)^2 \Big{)}\text{ d}x$, where $a<b$ and $a$ and $b$ are the $x$-values of the intersection points.$

calebwhitaker199 · Nov 11, 2015

Thank you for that, in this case would the boundaries of the integration be 2 and zero ? My answer turned out to be zero for the first question.

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InteGrand · Nov 11, 2015

calebwhitaker199 said:
Thank you for that, in this case would the boundaries of the integration be 2 and zero ? My answer turned out to be zero for the first question.

Sent from my iPad using Tapatalk

$For the first question, the formula in my previous post does not apply, since the region is not one where one of the given curves forms the ``upper boundary'' of the region, and the other the lower boundary. Here is a graph of the situation:$

http://www.graphsketch.com/?eqn1_co..._lines=1&line_width=4&image_w=850&image_h=525

$The region being rotated is that between the two curves plotted and the horizontal axis (between $x=0$ and 2).$

$Note that if we draw a dotted vertical line at $x=1$, this is the point where the two curves meet. By symmetry, rotating the region about the $x$-axis results in a volume that is \textsl{double} that of just rotated the area under the curve of $y=x^2$ (the blue curve) between $x=0$ and $x=1$.$

$This latter volume is$

$\begin{align*}\pi \int _0^1 \left( x^2\right)^2 \text{ d}x &=\pi \int _0^1 x^4 \text{ d}x \\ &= \pi\left[ \frac{x^5}{5}\right] _0 ^1 \\ &= \pi\left(\frac{1^5}{5}-\frac{0^5}{5}\right)\\ &= \frac{\pi}{5} .\end{align*}$

$So the required volume is double this, or $\frac{2\pi}{5}$. Hence the answer is \boxed{\frac{2\pi}{5}\text{ cubic units}}.$

Help with Integration questions (1 Viewer)

calebwhitaker199

New Member

InteGrand

Well-Known Member

calebwhitaker199

New Member

InteGrand

Well-Known Member

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