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Math Help - Calc volume / area between curve help:

  1. #1
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    Calc volume / area between curve help:

    Hey guys. I'm just looking for some quick math help. These 2 problems are about areas between curves and the volume methods of integration.

    1. Find area of region bounded by curves y = x + 5, x = y^2, y = -1, and y = 2.

    For #1, I got 16.5.

    2. Let R be the region bounded by the x-axis and the curves y = (2+5x^2) ^ (-1/2), x = 1, and x = 4. Find volume obtained by rotating R about y-axis.

    For some reason I keep getting a negative answer here. That can't be right for a volume problem. Also, another random note: it does say rotate around y-axis, but even if you did it around x-axis, do you get the same answer?
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  2. #2
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    Quote Originally Posted by XFactah416 View Post
    Hey guys. I'm just looking for some quick math help. These 2 problems are about areas between curves and the volume methods of integration.

    1. Find area of region bounded by curves y = x + 5, x = y^2, y = -1, and y = 2.

    For #1, I got 16.5.
    How did you get this answer?


    Quote Originally Posted by XFactah416 View Post
    2. Let R be the region bounded by the x-axis and the curves y = (2+5x^2) ^ (-1/2), x = 1, and x = 4. Find volume obtained by rotating R about y-axis.

     <br />
V = \pi\int_1^4 y^2~dx<br />
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  3. #3
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    Quote Originally Posted by pickslides View Post
    How did you get this answer?





     <br />
V = \pi\int_1^4 y^2~dx<br />
    For #1:

    I chose to use horizontal strips, so I converted y = x + 5 into x = y-5.

    Thus, I had an interval of -1 to 2, and it was [right curve - left curve] dy. My right curve was y^2, and my left curve was (y-5).

    After all the math, I got 16.5. I'm guessing it's incorrect?

    For #2, are you just saying to square (2+5x^2) ^ (-1/2) and then take the integral of that? I have to rotate around the y-axis in this problem. I was under the impression that when you rotate around the y axis, you must convert the equation into an x = y equation.
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    Quote Originally Posted by XFactah416 View Post
    For #1:

    I chose to use horizontal strips, so I converted y = x + 5 into x = y-5.

    Thus, I had an interval of -1 to 2, and it was [right curve - left curve] dy. My right curve was y^2, and my left curve was (y-5).

    After all the math, I got 16.5. I'm guessing it's incorrect?
    O.k, it really depends on the width of your horizontal strips, could be correct.

    Quote Originally Posted by XFactah416 View Post

    For #2, are you just saying to square (2+5x^2) ^ (-1/2) and then take the integral of that? I have to rotate around the y-axis in this problem. I was under the impression that when you rotate around the y axis, you must convert the equation into an x = y equation.
    yep, square it, integrate substituting in the terminals and multiply the result by pi.
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  5. #5
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    OK.

    Thank you! However, I'm guessing my theory is incorrect about rotation around y-axis means that we have to have an equation of x = y?
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  6. #6
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    Quote Originally Posted by XFactah416 View Post
    Hey guys. I'm just looking for some quick math help. These 2 problems are about areas between curves and the volume methods of integration.

    1. Find area of region bounded by curves y = x + 5, x = y^2, y = -1, and y = 2.

    For #1, I got 16.5. <<<<<<<< OK

    2. Let R be the region bounded by the x-axis and the curves y = (2+5x^2) ^ (-1/2), x = 1, and x = 4. Find volume obtained by rotating R about y-axis.

    For some reason I keep getting a negative answer here. That can't be right for a volume problem. Also, another random note: it does say rotate around y-axis, but even if you did it around x-axis, do you get the same answer?
    1. The general form of the volume which you get by rotation about the y-axis is:

    V_{roty} = \int_{f(a)}^{f(b)} \pi x^2 dy

    2. To calculate the volume it isn't necessary to determine x but x:

    y = \dfrac1{\sqrt{2+5x^2}}~\implies~x^2 = \dfrac1{5y^2} - \dfrac25

    If y = f(x) you'll get f(1) =  \frac1{\sqrt{7}} and f(4) = \frac1{\sqrt{22}}

    The complete equation to calculate the volume is

    V_{roty}=\int_{\tfrac1{\sqrt{22}}}^{\tfrac1{\sqrt{  7}}} \left( \pi \cdot \left(\dfrac1{5y^2} - \dfrac25 \right) \right) dy

    3. I've got an approximate value of the volume of 1.07765

    4. The volume of rotation at rotation about the x-axis is calculated by:

    V_{rotx} = \pi \int_a^b y^2 dx

    In genaral the volumes V_{rotx} and V_{roty} are unequal.
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    Quote Originally Posted by XFactah416 View Post
    Hey guys. I'm just looking for some quick math help. These 2 problems are about areas between curves and the volume methods of integration.

    1. Find area of region bounded by curves y = x + 5, x = y^2, y = -1, and y = 2.
    What region are you talking about? The line y= x+ 5 does NOT intersect the curve x= y^2 between y= -1 and y= 2.

    For #1, I got 16.5.

    2. Let R be the region bounded by the x-axis and the curves y = (2+5x^2) ^ (-1/2), x = 1, and x = 4. Find volume obtained by rotating R about y-axis.

    For some reason I keep getting a negative answer here. That can't be right for a volume problem. Also, another random note: it does say rotate around y-axis, but even if you did it around x-axis, do you get the same answer?[/QUOTE]
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    What region are you talking about? The line y= x+ 5 does NOT intersect the curve x= y^2 between y= -1 and y= 2.

    ....
    When I checked the result I assumed that the region was meant which is coloured in grey. And then the posted result is OK.
    Attached Thumbnails Attached Thumbnails Calc volume / area between curve help:-area_wurz_gerade.png  
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