Here's the question you clicked on:
Moyo30
Find the volume of the torus (doughnut) formed by rotating the circle x2 + (y − 2)2 = 1 about the x-axis.
Hint: \[\int\limits_{-1}^{1}\sqrt{1-x ^{2}}=\frac{ \pi }{ 2 }\]
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Do you know the formula for the volume of solid rotating \(y=f(x)\) about \(x\)-axis?
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\[\pi \int\limits_{a}^{b}f(x)dx\]
|dw:1357858535828:dw| Subtract the right volume from the left. Got it?
|dw:1357858823945:dw| It is like subtracting this areas.
\(V=\pi\int_{-1}^1(\sqrt{1-x^2}+2)^2dx-\pi\int_{-1}^1(-\sqrt{1-x^2}+2)^2dx\)
If you don't get it now just sit and think about it very thoroughly.
|dw:1357858979556:dw| the graph of x^2+(y-2)^2=1 is something like that how do we get to the penultimate drawing
i kinda get it you reflect a portion of the circle in the x axis and find the region bounded by the two arcs|dw:1357859345394:dw|
if you try to write y depending on x, you will get 2 functions: \(y=\sqrt{1-x^2}+2\) and \(y=-\sqrt{1-x^2}+2\). Now think what will be if you rotate them about x-axes. And then use your imagination to unterstand how to get the needed volume. Really the plot is:|dw:1357859397346:dw|
But if you are lazy and dishonest student you can just use the formulas I've written above.
Have you understood something?
yes it is basically a volume but it complicated by the fact that the function that we get after making y the subject of the formula is composed of two equations so we add the two. right?
It is very hard to tell if you got something or not. I don't know your level of knowledge. Tell me honestly: can you find the area between \(y=x^2\) and \(y=1-x^2\)?
subtract x^2 from 1-x^2 and integrate
juss realised my prob no squaring
integrate, limits the intersection of the lines
Yes. The similar situation is here. But now we have a volume to subtract. Hope, you got something, if not - very bad.
so basically in this prob they ask us to find the volume rather than the area
can you pls draw the resultant volume if it is not too diff
Yes. The volume of the doughnut.|dw:1357861006645:dw|