My Math Forum Washer Method

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 May 5th, 2014, 11:28 AM #1 Senior Member   Joined: Dec 2013 Posts: 137 Thanks: 4 Washer Method find the volume V generated by rotating the given region about the specified line. a = 10 b = 3 here is a picture Gyazo - 94e1d3eeb1acbbeb8923017cc2e93828.png So here is what I did: I drew another square cell to the left w/coordinates (0,0), (-1,0),(-1,10),(0,10) next I drew the region in the new cell (symmetry is clear here so this part was not hard or anything) next, I notice the problem is telling me to rotate the given region around the line it gives (which just so happens to be the Y-Axis). Thus, when doing washer method I know the cross-section is going to be perpendicular to the axis being revolved around. So, I now see I must solve any given equations for Y (instead of x) doing so I got x = (y/10)^1/3 My first question to the forum is ... DID I DO THAT RIGHT? Next, I know I use the formula of Integrating from 0 to 10 of Pi(R^2 - r ^2)dy and here I am stuck. I wish to understand what R^2 = and what little r ^2 = I want to go (1 - (y/10)^1/3)^2 for Big R but I feel here is where I am getting off the path, etc. Thanks forum for any help you guys can do for me on this problem. I know it is essentially the concept I am trying to master so I am doing a lot of the disk and also the washer method problems to get better as well as the cylindrical shell method problems.
 May 5th, 2014, 01:55 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Your equation for the curve is fine. But the curve is the inner radiius of your washer is the curve and the outer radius is the line $x=1$. Thanks from mr. jenkins
 May 8th, 2014, 04:33 PM #3 Senior Member   Joined: Dec 2013 Posts: 137 Thanks: 4 @V8 still struggling bud here is what I got R = pi[(1-(y^(1/3)/10)]^2 r = pi[(1-(y/10)]^2 = A(y) = pi[1-(2y^(1/3)/10) + (y^(2/3))/10] - [1-2y/10 + y^(2)/10] = A(y) = pi[ -2y^(1/3)/10) + y^(2/3)/10) + (2y/10) - y^(2)/10)] dy = V = pi [-3y^4/3)/20) + 3y^(5/3)/50) + y^(2)/10) - y^(3)/30)] from 0 to 10 = now if I am right so far...I have questions ... #1 Can I go ahead and get common denominators so that I can then combine the terms -3y^4/3 /20 and 3y^5/3 / 50 because if so I wonder if the powers would wash out ...elsewise raising 10 to the 4/3 and 10 to the 5/3 seems to give a decimal answer and this is where/why I feel I have done something wrong. I think I will stop here V8 and wait to hear back from you before I venture into the weeds any further. Thank you and I will eagerly await your reply. thanks.
 May 13th, 2014, 07:17 PM #4 Senior Member   Joined: Dec 2013 Posts: 137 Thanks: 4 For the life of me ... I cannot crack this problem and really need some help with it to see where I am wrong
 May 13th, 2014, 07:48 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra I'll work through it. I think you have most of it sorted though. OK, so we want to rotate an area around the y-axis. So it'll be easiest to write all equations as $x = f(y)$. The outer curve is $x = 1$, the inner curve is given as $y = 10x^3$ so $$x = \left(\frac{y}{10}\right)^\frac{1}{3}$$, so in your terms $$R - r = 1 - \left(\frac{y}{10}\right)^\frac{1}{3}$$ We are going to integrate with respect to $y$ between 0 and a = 10. So let's set up the integral to find the volume $V$ \begin{align*} V &= \int_0^{10}{\pi\left(1^2 - \left(\left(\frac{y}{10}\right)^\frac{1}{3}\right) ^2 \right)}dy \\ &= \pi\int_0^{10}{1 - \frac{y^\frac{2}{3}}{10^\frac{2}{3}}}dy \\ &= \pi \left. \left(y - \frac{1}{10^\frac{2}{3}} \frac{3}{5} y^\frac{5}{3} \right) \right|_0^{10} \\ &= \pi \left( \left( 10 - \frac{1}{10^\frac{2}{3}} \frac{3}{5} {10}^\frac{5}{3} \right) - \cancel{\left( 0 - \frac{1}{10^\frac{2}{3}} \frac{3}{5} {0}^\frac{5}{3} \right)} \right) \\ &= \pi \left( 10 - \frac{3}{5} 10^{\frac{5}{3}-\frac{2}{3}} \right) \\ &= \pi \left( 10 - \frac{3}{5} 10 \right) \\ &= \frac{2}{5} 10 \pi \\ &= 4\pi \\ \end{align*}
 May 13th, 2014, 08:21 PM #6 Senior Member   Joined: Dec 2013 Posts: 137 Thanks: 4 V8 that is so awesome. so easy for you haha I was also so close. the 4pi is correct also, via verified. well...wife is kicking me and my math books out of the den. thank you thank you so much and of course i will keep this one for practice
 May 14th, 2014, 07:33 PM #7 Senior Member   Joined: Dec 2013 Posts: 137 Thanks: 4 yes ty again v8 i am doing other similar problems and grasping the concept of the inner radius being the function or curve and outer radius being the given x value at least this seems to be the case for rotating about the y axis math is funny v8 so much of the time it makes no sense whatsoever, until it makes sense and then it is just childs play!

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