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Thread: 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 $\displaystyle y = 4 - x^2$ and a straight line $\displaystyle y=7 - 4x$ which is tangent to the curve at the point $\displaystyle (1 , 3)$


    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.

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

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

    There you are!

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

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

    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.

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

    because it rotated $\displaystyle 360^\circ$about 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:

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

    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)

    $\displaystyle \text{Volume} =\displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2 - \left(\sqrt{4-y}\right)^2\, dy$
    Last edited by Unknown008; Oct 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 $\displaystyle y = 4-x^2$

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

    not use $\displaystyle \pi \int^4_3 \left(\sqrt{4-y}\right)^2 dy$?
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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:

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

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

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

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

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

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

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

    $\displaystyle (7-y)^2 = 49 - 14y + y^2$

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

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

    and how to integrate $\displaystyle \displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2$
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  13. #13
    MHF Contributor Unknown008's Avatar
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    $\displaystyle \displaystyle \pi \int^7_3 \left(\dfrac{7-y}{4}\right)^2 = \pi\int^7_3 \dfrac{49-14y+y^2}{16}\ dy= \dfrac{\pi}{16} \int^7_3 49-14y+y^2 \ dy$
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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 :

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

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

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

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

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

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

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