anonymous
  • anonymous
Let R be the region in the first quadrant bounded by the y-axis, the curve y=sqr(1-x^2)and the line y=xsqr(3) Find the volume of the solid obtained by revolving R about the x-axis.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ganeshie8
  • ganeshie8
As a start, sketch the given curves
ganeshie8
  • ganeshie8
|dw:1450110279849:dw|
anonymous
  • anonymous
yes

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ganeshie8
  • ganeshie8
familar with discs method ?
anonymous
  • anonymous
how do I know to use discs?
ganeshie8
  • ganeshie8
you can use either method
anonymous
  • anonymous
but discs would be best
ganeshie8
  • ganeshie8
discs or shell
ganeshie8
  • ganeshie8
Yes, how do you know discs would be the best ?
anonymous
  • anonymous
because it makes it more even to split
ganeshie8
  • ganeshie8
discs is indeed the best here
ganeshie8
  • ganeshie8
consider a small volume element
ganeshie8
  • ganeshie8
|dw:1450111537556:dw|
ganeshie8
  • ganeshie8
Volume of that thin shell of thickness \(\Delta x\) is given by : \[\Delta V = A*\Delta x = (\pi y_1^2 - \pi y_2^2)*\Delta x\]
ganeshie8
  • ganeshie8
plugin \(y_1 = \sqrt{1-x^2}\) and \(y_2 = x\sqrt{3}\)
anonymous
  • anonymous
okay!
ganeshie8
  • ganeshie8
\[\Delta V = (\pi (1-x^2) - 3\pi x^2 )*\Delta x\] In the limit case, it becomes \[dV = (\pi (1-x^2) - 3\pi x^2 )*dx\]
ganeshie8
  • ganeshie8
Integrate that from \(x=0\) to \(x=1/2\) to get the desired volume
ganeshie8
  • ganeshie8
\[V = \int\limits_{0}^{1/2}(\pi (1-x^2) - 3\pi x^2 )\,dx\]
anonymous
  • anonymous
why to 1/2?
ganeshie8
  • ganeshie8
good question, that is the reason i have asked you to sketch the curves first
ganeshie8
  • ganeshie8
look at the curves where do they intersect ?
ganeshie8
  • ganeshie8
|dw:1450112179986:dw|
anonymous
  • anonymous
okay why did the square root signs go away?
ganeshie8
  • ganeshie8
good question, Volume of that thin shell of thickness \(\Delta x\) is given by : \[\Delta V = A*\Delta x = (\pi y_1^2 - \pi y_2^2)*\Delta x\]
ganeshie8
  • ganeshie8
plugin \(y_1 = \sqrt{1-x^2}\) and \(y_2 = x\sqrt{3}\) what do you get ?
anonymous
  • anonymous
oh! hahahaha
anonymous
  • anonymous
this may be a stupid question but when you integrate, do you distribute pi first?
anonymous
  • anonymous
\[\pi \left( x-\frac{ x^{3} }{ 3 }-x ^{3} \right)\]
anonymous
  • anonymous
this is what i have so far
ganeshie8
  • ganeshie8
looks good, next apply the bounds 0 -> 1/2
anonymous
  • anonymous
5/48?
ganeshie8
  • ganeshie8
wolfram give pi/3 http://www.wolframalpha.com/input/?i=+%5Cint%5Climits_%7B0%7D%5E%7B1%2F2%7D%28%5Cpi+%281-x%5E2%29+-+3%5Cpi+x%5E2+%29%5C%2Cdx

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