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Math Help - problem in region bounded by a curve and a straight line

  1. #1
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    problem in region bounded by a curve and a straight line

    Diagram shows a curve y = 4 - x^2 and a straight line y=7 - 4x which is tangent to the curve at the point (1 , 3)


    a)calculate the area of the shaded region.
    Attached Thumbnails Attached Thumbnails problem in region bounded by a curve and a straight line-graph.png  
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  2. #2
    MHF Contributor Unknown008's Avatar
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    There is no diagram...

    I guess it's the area between the line and the curve. Use integration and the area of a trapezium.

    \text{Area of trapezium} - \int^1_0 4 - x^2 dx

    \frac12(7 + 3)(1) - \int^1_0 4 - x^2 dx

    There you are!

    EdIT: Ah, that's what I thought the diagram was.
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  3. #3
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    How about the volume of revolution,in term of \pi when the shaded region is rotated 360^\circ about the y-axis?
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Well, you use the formula:

    Volume = \pi \int^1_0 y^2 dx

    Here, it'll be preferable to use the equation of the straight line itself instead of a shortcut like I did for the trapezium above.

    Volume = \pi \int^1_0 (7-4x)^2 - (4-x^2)^2 dx
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  5. #5
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    are u sure use the formula Volume = \pi \int^1_0 y^2 dx not
    Volume = \pi \int^1_0 x^2 dy?

    because it rotated 360^\circabout the y-axis
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Oh, I didn't notice you were rotating around the y-axis my bad...

    Right, we use the formula:

    \text{Volume} =\displaystyle \pi \int^b_a x^2\, dy

    Since you are rotating through the y-axis, you need the y coordinates.

    So, we get:

    y = 7 - 4x
    x = (7 - y)/4

    y = 4 - x^2
    x = \sqrt(4 - y)

    \text{Volume} =\displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2 - \left(\sqrt{4-y}\right)^2\, dy
    Last edited by Unknown008; October 17th 2010 at 06:44 AM. Reason: added some more details, and found how to make a big integral sign XD
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  7. #7
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    sir,at the curve y = 4-x^2

    why we use \pi \int^7_3 \left(\sqrt{4-y}\right)^2 dy?

    not use \pi \int^4_3  \left(\sqrt{4-y}\right)^2 dy?
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  8. #8
    MHF Contributor Unknown008's Avatar
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    You're right again >_<

    I must be tired... sorry for that
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  9. #9
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    oh..its ok sir..u need to rest..thanks for the help sir!
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  10. #10
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    hi..check this:

    \text{Volume} =\displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2-\pi \int^4_3 \left(\sqrt{4-y}\right)^2 dy

    =\displaystyle \pi \int^7_3 \left(\dfrac{49-y^2}{16}\right) - \pi \int^4_3 \left(\sqrt{4-y}\right)^2 dy

    = \pi \left[\frac{49y - y^3}{16(3)}\right]^7_3 - \pi\left[4y - y^2\right]^4_3

    = \left [\frac{49(7) - 7^3}{48} - \frac{49(3) - 3^3}{48} \right] - \pi\left[4(4) - 4^2 - 4(3) - 3^2\right]

    = \pi \left( -\frac{5}{2} \right) - \pi ( -21)

    = \frac{37}{2} \pi unit^3

    am i right sir?
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  11. #11
    MHF Contributor Unknown008's Avatar
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    No

    (7-y)^2 = 49 - 14y + y^2

    And \int 4-y\ dy = 4y - \dfrac{y^2}{2} + c
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  12. #12
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    urm..

    why you different (7-y)^2 = 49 - 14y + y^2?

    and how to integrate \displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2
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  13. #13
    MHF Contributor Unknown008's Avatar
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    \displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2 = \pi\int^7_3 \dfrac{49-14y+y^2}{16}\ dy= \dfrac{\pi}{16} \int^7_3 49-14y+y^2 \ dy
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  14. #14
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    ok thanks..i try do it now..later,,can u check my answer?
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  15. #15
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    i get :

    \pi\int^7_3 \dfrac{49-14y+y^2}{16}\ dy - \pi \int^4_3 \left(\sqrt{4-y}\right)^2 dy

    = \dfrac{\pi}{16} \left[49y-\frac{14y^2}{2}+ \frac{y^3}{3}\right]^7_3 - \pi\left[4y - \frac{y^2}{2} \right]^4_3

    \dfrac{\pi}{16} \left[49(7) - \frac{14(7)^2}{2} + \frac{7^3}{3}\right] - \left[49(3) - \frac{14(3)^2}{2} + \frac{3^3}{3} \right] - \pi\left[4(4)  - \frac{4^2}{2}- 4(3) - \frac{3^2}{2} \right]

    = \dfrac{\pi}{16} \left[\frac{343}{3} - 147 + 63 - 9\right] - \pi\left[8 - \frac{15}{2}\right]

    = \dfrac{\pi}{16} \left[\frac{64}{3}\right] - \pi\left[\frac{1}{2}\right]

    = \frac{4}{3}\pi - \frac{1}{2}\pi

    = \frac{5}{6} unit^3
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