anonymous
  • anonymous
rotation of a bounded area question i can't solve Write the integral in one variable to find the volume of the solid obtained by rotating the first-quadrant region bounded by y = 0.5x^2 and y = x about the line x = 5.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
my trouble is mainly with the fact that the rotation is done around x=5 and not the x axis
anonymous
  • anonymous
im sure something changes but not sure what
anonymous
  • anonymous
Do you prefer the washer or shell method?

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anonymous
  • anonymous
well i don't know the shell method so washer i suppose
anonymous
  • anonymous
Observe that the dimensions of the washer that you'll be interested in (i.e. radius) will have to be written in terms of \(y\). Let's draw the region first: |dw:1439519332453:dw|
anonymous
  • anonymous
so does that mean we will have to integrate with sespect to y?
anonymous
  • anonymous
Yes! Since \(y=\dfrac{1}{2}x^2\), you have \(x=\sqrt{2y}\) (taking the positive root since we're in the first quadrant). Here's a washer associated with an matching of points \((x,y)=(\sqrt{2y},y)\) and \((x,y)=(y,y)\) on the boundary of the region: |dw:1439519657597:dw|
anonymous
  • anonymous
(As you'll no doubt learn soon enough, the shell method allows you to integrate with respect to \(x\) instead.)
anonymous
  • anonymous
Correction: the inner radius should be \(5-\color{red}{\sqrt{2y}}\)
anonymous
  • anonymous
The volume of the solid is the infinite sum of the areas of these infinitely thin washers. The area of a washer is essentially the difference of the areas of circles. Take some circle with radius \(OR\) and put some smaller circle with radius \(IR\) at the larger one's center. The area of the large circle is \(\pi(OR)^2\), and that of the smaller circle is \(\pi(IR)^2\). Remove the smaller circle from the larger, and you're left with the washer shape, which has area \(\pi((OR)^2-(IR)^2)\). So, to add up all these areas, you take the integral. It'll look something like this (I'll leave the details to you): \[V=\int_a^b \pi((OR)^2-(IR)^2)\,dy\]
anonymous
  • anonymous
okay sorry lost internet connection , back now give me a second to read all that you posted
anonymous
  • anonymous
okay so tell me if this is way off \[V = \pi \int\limits_{0}^{2}(y^2)-({2y}) dy\]
anonymous
  • anonymous
It's off by little bit. Your setup would be right if you were revolving about the \(y\)-axis (the line \(x=0\)).
anonymous
  • anonymous
okay so do i just add 5 to each expression or something?
anonymous
  • anonymous
Sort of... if you take another look at that sketch, you'll notice that I lay out the lengths of a few distances. Let me remove some of that clutter: |dw:1439520860137:dw| Notice that the two smaller distances must add up to \(5\). That is, \[y+OR=5\] This means \(OR=5-y\), not just \(y\).
anonymous
  • anonymous
Similarly, you can figure out what the inner radius would be by breaking down these distances: |dw:1439521024630:dw| You have \(\sqrt{2y}+IR=5\), so \(IR=5-\sqrt{2y}\).
anonymous
  • anonymous
okay thats interesting let me give it another shot \[V = \pi \int\limits_{0}^{2} (-x-5)^2 - (5-\sqrt{2y}) dy\]
anonymous
  • anonymous
woops forgot to square the last part
anonymous
  • anonymous
I think you meant \(5-x\), or \(-x+5\) for the first term? Squared, of course.
anonymous
  • anonymous
\(y\), not \(x\)**
anonymous
  • anonymous
yes yes sorry about that
anonymous
  • anonymous
im getting 5.333 , does that look right to you?
anonymous
  • anonymous
Yup
anonymous
  • anonymous
times \(\pi\), right?
anonymous
  • anonymous
oh yea forgot :)
anonymous
  • anonymous
It doesn't look like you actually had to do the computation, but maybe those instructions were just left out? Anyway, setting up these types of integrals is the hardest part.
anonymous
  • anonymous
(usually)
anonymous
  • anonymous
thanks for the help ! :D
anonymous
  • anonymous
You're welcome!

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