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 February 16th, 2011, 11:46 PM #1 Newbie   Joined: Jan 2011 Posts: 17 Thanks: 0 Integration. About -ve area and stationary value. Greetings to all. Okay, I have 2 simple questions about integration. They are summarized in topic title. 1) This is the figure: It shows a curve y=x^2 and the line y=3+2x. Point A is (3,9) and point B is (-1,1). We have to calculate the area of the region enclosed between the curve and the line segment AB. I have used the formula A=$integration$ (y2-y1) dx. Correct me if the formula is wrong. I've used the limits 3 and -1. Now, I have solved this and obtained the answer, but the problem is that the Area I'm getting is coming in negative. The answer to this question is 32/3 which is what I'm getting but only that it is -32/3. Is this something to do with the figure or have I done something wrong? 2) This is the second part of a question. First part is done. We were asked to show the Volume, V cm^3, was given by: V=12$pi$r^2-2$pi$r^3 [This is correct BTW]. Okay, now the second part of the question says that Given that r varies, find the stationary value of V. What I did here is, because the stat.value says that dV/dx = 0, I used: 0(dV/dx)=12$pi$r^2-2$pi$r^3. 0=24$pi$r-6$pi$r^2. After this, I don't know what else to do. The answer is coming as 64$pi$. But how is this coming? Any help? :
 February 17th, 2011, 12:10 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Integration. About -ve area and stationary value. Question 1: $A=\int_{-1}\,^{3}3+2x-x^2\,dx$ $A=$3x+x^2-\frac{1}{3}x^3$_{-1}^{3}=$$3(3)+(3)^2-\frac{1}{3}(3)^3$$-$$3(-1)+(-1)^2-\frac{1}{3}(-1)^3$$=$ $(9)-$$-\frac{5}{3}$$=\frac{32}{3}$ Question 2: $V=12\pi r^2-2\pi r^3$ $\frac{dV}{dr}=24\pi r-6\pi r^2=0$ Divide through by $-6\pi$ : $r^2-4r=0$ Factor: $r(r-4)=0$ Since $\frac{d^2V}{dr^2}=24\pi$ at r = 0, and $\frac{d^2V}{dr^2}=-24\pi$ at r = 4 we determine that V has local minimum at r = 0 and a local maximum at r = 4. Thus $V(4)=12\pi (4)^2-2\pi \(4)^3=64\pi$
 February 17th, 2011, 01:11 AM #3 Newbie   Joined: Jan 2011 Posts: 17 Thanks: 0 Re: Integration. About -ve area and stationary value. Woah! Can't believe I messed up the equation's arrangement. Gotta remember that about finding area in integration. About the second question, just one thing. You got r(r-4)=0 which would give us either r=0 or r=4. If we use r=0 in the original equation, we get left with 0. When we use r=4 in the original equation, we get 64.pi. So, we only take this value as the answer. But what if it was say (r-3)(r-4)=0? We'd have r=4 and r=3. So if plugged in r=3, we'd get : 12.pi(3)^2 - 2.pi(3)^3. This would give us 54.pi. Would we have 2 answers then? How would we write it? A little more help with the last one and this problem is done. Thanks for the help.
 February 17th, 2011, 07:00 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Integration. About -ve area and stationary value. It would depend on what type of extrema we were interested in finding. In this problem we were presumably looking for 0 < r < 6 that maximized V. You basically look at the behavior of the function over the domain of the independent variable and choose which extrema meets the criteria, if any at all. If maximizing or minimizing, you also need to look at the end-points to determine if those values are greater than or less than the local extrema.
 February 17th, 2011, 09:12 AM #5 Newbie   Joined: Jan 2011 Posts: 17 Thanks: 0 Re: Integration. About -ve area and stationary value. Thanks for the explanation. Guess this one is solved.

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