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anonymous
 one year ago
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 firstquadrant region bounded by y = 0.5x^2 and y = x about the line x = 5.
anonymous
 one year ago
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 firstquadrant region bounded by y = 0.5x^2 and y = x about the line x = 5.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my trouble is mainly with the fact that the rotation is done around x=5 and not the x axis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im sure something changes but not sure what

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you prefer the washer or shell method?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well i don't know the shell method so washer i suppose

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Observe 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
 one year ago
Best ResponseYou've already chosen the best response.0so does that mean we will have to integrate with sespect to y?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! 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
 one year ago
Best ResponseYou've already chosen the best response.0(As you'll no doubt learn soon enough, the shell method allows you to integrate with respect to \(x\) instead.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correction: the inner radius should be \(5\color{red}{\sqrt{2y}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0okay sorry lost internet connection , back now give me a second to read all that you posted

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so tell me if this is way off \[V = \pi \int\limits_{0}^{2}(y^2)({2y}) dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's off by little bit. Your setup would be right if you were revolving about the \(y\)axis (the line \(x=0\)).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so do i just add 5 to each expression or something?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sort 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 \(OR=5y\), not just \(y\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Similarly, 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
 one year ago
Best ResponseYou've already chosen the best response.0okay thats interesting let me give it another shot \[V = \pi \int\limits_{0}^{2} (x5)^2  (5\sqrt{2y}) dy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0woops forgot to square the last part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you meant \(5x\), or \(x+5\) for the first term? Squared, of course.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes yes sorry about that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im getting 5.333 , does that look right to you?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0times \(\pi\), right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.0thanks for the help ! :D
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