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Volume Problems

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data

    1. The base of a solid S is the shaded region in the xy-plane enclosed by the x-axis, the y-axis and the graph of y=1-sinx. For each x, the cross section of S perpendicular to the x-axis at the point (x,0) is an isosceles right triangle whose hypotenuse lies in the xy-plane.

    (a) Find the area of the triangle as a function of x.

    (b) Find the volume of S.

    2. An oil storage has the shape obtained by revolving the curve y=9*x^4/625 from x=0 to x=5 about the y-axis, where x and y are measured in feet. Oil flows into the tank at the constant rate of 8 cubic feet/min.

    (a) find the volume of the tank. Indicate units of measure.

    (b) To the nearest minute, how long would it take to fill the tank if the tank was empty initially ?

    (c) Let h be the depth ,in feet, of oil in the tank. How fast is the depth of the oil in the tank increasing when h=4 ? Indicated units of measure.

    2. Relevant equations

    3. The attempt at a solution

    For number one, I found that the lenght of the triangle is 1-sinx. so is the area just (1-sinx)^2/2 ??

    I am totally lost for others. Really URGENT
  2. jcsd
  3. Nov 11, 2008 #2


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    If this is URGENT, you'd better get busy. You've already got a problem with 1a). 1-sin(x) is the length of the hypotenuse of the right triangle. Draw one. What's the area? The sooner you present an attempt at a solution for the other parts, the sooner you will get help.
  4. Nov 11, 2008 #3
    so side is 1-sin(x)/ sqroot(2) ? Thus, the area will be (1-sin(x)^2/4 ?

    For part b, will it be the integral from 0 to 1.571 ??
  5. Nov 11, 2008 #4


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    Yes, that's the area. You mean the integral of the area, right? Sure. But I think they will want you to give an exact value of the limit instead of the approximation 1.571. Did you get that from a graphing calculator or something?
  6. Nov 11, 2008 #5
    Yeah, I did. Which can be a more accurate way to find out the x interception ?
  7. Nov 11, 2008 #6


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    The x intercept is where y=0. Solve y=1-sin(x)=0. Look at a graph of sin.
  8. Nov 11, 2008 #7
    I got a really weird thing from my TI-89. Should I use the approximation instead then ?
  9. Nov 11, 2008 #8
    Stupid me it is pi/2.

    I am really lost at the 2nd problem. Can you help me ?
  10. Nov 11, 2008 #9
    Something that might help you:
    Area from a-->b = integral from a-->b of f(x) dx
    Volume from a-->b = (pi)(integral from a-->b of f(x)squared dx)
    So for #1 find the integral of: 1-sinx dx (from 0 to pi/2)
    See what you get!
  11. Nov 11, 2008 #10
    I got about 0.089.

    Ok. I got number 1.

    But number 2 I am kinda lost. Should I use shell or dish method here ? If I use dish, I have limits of integration from 0 to 5 but what can be the integrand ?

    If I use shell. I will have x= (y*625/9)^(1/4). Thus, my formula will be something like

    pi* integral ( x^2, evaluation from 0 to 9) ?? Is it true ? Which way should I follow ?
  12. Nov 11, 2008 #11
    Ok first make sure if you use your calculator that it is in radians... because that is not what I got for the area of #1. By dish method do you mean disk lol? Anyways, since you revolve around the y axis, you have to put your function in terms of x =, now i dont know if what you have is right though.
  13. Nov 11, 2008 #12
    What did you get ? That's what I got in radian mode. Did you square (1-sin(x)) ? and then divide the integral by 4 ?

    What do you mean ? I meant disk, sorry.

    I did put function in term of x like above. is that right ?
  14. Nov 11, 2008 #13
    Sorry got back from lunch,
    Why would i square 1-sin(x) and then divide the integral by 4?? do you know how to get the integral of : f(x) = 1-sin(x) ??
    Well is the equation: (9(x^4))/625 OR (9)(x^(4/625)) ??
  15. Nov 11, 2008 #14


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    You would do those things because that how you get the cross sectional area to integrate. nns91 is doing it correctly, the answer is about 0.089. I don't think you read the problem completely. But, again, nns91, I think you want to express the answer exactly instead of as an approximation. Can't you do the integral without doing it numerically?
  16. Nov 11, 2008 #15
    I could but I think my teacher prefer approximation. Thanks for your comment.

    The problem is (9*x^4)/625.

    So how can I do the 2nd problem ?
  17. Nov 11, 2008 #16


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    Do the next one like you did the first one. Sure, use disks. Given a value of y, what's the radius of the disk? What's the area of the disk? Now integrate area dy to get the volume.
  18. Nov 11, 2008 #17
    Radius will be x=((625*y)/9)^1/4 so area will be pi*x^2 ??? Is that true ?

    I am confused about part c too.

    So v= pi*r^2 * h right ? What's next ? I too the derivative but it led to dr/dt which I don't know ?
  19. Nov 11, 2008 #18


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    Right. Area is pi*x^2=pi*((y*625)/9)^(1/4))^2. But you don't just multiply area times height, the area isn't constant. You have to INTEGRATE area*dy. Because you know part c) is coming don't just find the volume for h=5. Find the volume for a general value of h.
  20. Nov 11, 2008 #19
    so how can I find the limit of integral is that 0 to 9 ?

    v=pi*r^2*h right ? so I just take the derivative. and gt dv/dt = 2pi*r*h*dr/dt + 2pi*r*dh/dt.

    I have pretty much everything except dr/dt and r. How can I find those variables ?
  21. Nov 11, 2008 #20


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    Maybe I didn't emphasize that V=pi*r^2*h is WRONG. You have to INTEGRATE to find the volume. Go back to part a).
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