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    Double integral for x limit, 0 to 1 and y limit x to root(x) with respect to dydx

    Doing the inner integral gives 4(x^2)y + 4y^3/3, sub in the limits we get

    (4x^5/2 + 4(x^3/2)/3) - (4x^3 + 4x^3/3)

    Integrate again, we get (8(x^7/2)/7 + 8(x^5/2)/15) - (x^4 + 4x^4/12)

    We forget the minus as that is related to the zero limit so focus on the upper limit 1 and yields an area of 88/105 and the answer is supposed to be 36/105. Am I wrong or is the sheet wrong?

    Thanks in advance.
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    (Original post by boromir9111)
    Double integral for x limit, 0 to 1 and y limit x to root(x) with respect to dydx

    Doing the inner integral gives 4(x^2)y + 4y^3/3, sub in the limits we get

    (4x^5/2 + 4(x^3/2)/3) - (4x^3 + 4x^3/3)

    Integrate again, we get (8(x^7/2)/7 + 8(x^5/2)/15) - (x^4 + 4x^4/12)

    We forget the minus as that is related to the zero limit so focus on the upper limit 1 and yields an area of 88/105 and the answer is supposed to be 36/105. Am I wrong or is the sheet wrong?

    Thanks in advance.
     \displaystyle \left[ \dfrac{8}{7}x^{\frac{7}{2}} + \dfrac{8}{15}x^{\frac{5}{2}} - \dfrac{4}{3}x^4 \right]^1_0 = \dfrac{8}{7} + \dfrac{8}{15} - \dfrac{4}{3} = \dfrac{36}{105} =\dfrac{12}{35}
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    (Original post by Clarity Incognito)
     \displaystyle \left[ \dfrac{8}{7}x^{\frac{7}{2}} + \dfrac{8}{15}x^{\frac{5}{2}} - \dfrac{4}{3}x^4 \right]^1_0 = \dfrac{8}{7} + \dfrac{8}{15} - \dfrac{4}{3} = \dfrac{36}{105} =\dfrac{12}{35}
    1^5/2 = 1/2, 1^7/2 = 1/2 thus giving me a fraction different to yours. How did you get 8/7, 8/15?
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    (Original post by boromir9111)
    1^5/2 = 1/2, 1^7/2 = 1/2 thus giving me a fraction different to yours. How did you get 8/7, 8/15?
    :erm:

    1^{5/2}=\sqrt{1^5}=\sqrt{1}=1. :yep:
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    (Original post by james.h)
    :erm:

    1^{5/2}=\sqrt{1^5}=\sqrt{1}=1. :yep:
    Cheers mate.
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    (Original post by boromir9111)
    1^5/2 = 1/2, 1^7/2 = 1/2 thus giving me a fraction different to yours. How did you get 8/7, 8/15?
    :teehee:

    (brackets, brackets, brackets!)
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    (Original post by Clarity Incognito)
    :teehee:

    (brackets, brackets, brackets!)
    Oh dear me. What a silly error to make lol. Cheers for that mate, much appreciated
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    "Evaluate the double integral for x, from 1 to e and y, from 1 to e

    y/x . dxdy

    Reverse the order of integration and re-evaluate."

    So from dxdy I can get the answer they want on the sheet which is, (1+e^2)/4 but the order reversed, dydx, I get (e^2 - 3)/4

    So, limits are for y, from 1 to x and for x, 1 to e. Is this correct?
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    (Original post by boromir9111)
    "Evaluate the double integral for x, from 1 to e and y, from 1 to y

    y/x . dxdy

    Reverse the order of integration and re-evaluate."

    So from dxdy I can get the answer they want on the sheet which is, (1+e^2)/4 but the order reversed, dydx, I get (e^2 - 3)/4

    So, limits are for y, from 1 to x and for x, 1 to e. Is this correct?
    Draw a diagram for the function, and it'll become clear where to integrate across.
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    (Original post by Don John)
    Draw a diagram for the function, and it'll become clear where to integrate across.
    Yeah, I did draw a diagram but I am fishy on it tbh, is my limits wrong?
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    I've just read you've put integrating y from 1 to y, this makes no sense, do you mean integrating x from 1 to y?
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    (Original post by Don John)
    I've just read you've put integrating y from 1 to y, this makes no sense, do you mean integrating x from 1 to y?
    Oh my bad, I meant 1 to e for my limits as y. Will change that now. That is what is given, I have to re-evaluate to integrate dydx.
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    (Original post by boromir9111)
    Oh my bad, I meant 1 to e for my limits as y. Will change that now. That is what is given, I have to re-evaluate to integrate dydx.
    Here's a diagram. Does this help?
    Attached Images
     
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    (Original post by Don John)
    Here's a diagram. Does this help?
    Actually that image worked perfect mate. I totally understand it now!!! thanks for that. + repped.
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    double integral for y, from 0 to pi/2 and x, from 0 to y

     \cos(2y) \times \sqrt{1 - k^{2}sin^{2}x}    dxdy

    k is a constant.

    So, (1-k^2*sin^2(x))^1/2. How do I integrate this?
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    (Original post by boromir9111)
    double integral for y, from 0 to pi/2 and x, from 0 to y

     \cos(2y) \times \sqrt{1 - k^{2}sin^{2}x}    dxdy

    k is a constant.

    So, (1-k^2*sin^2(x))^1/2. How do I integrate this?
    A substitution of u=(1-k^2sin^2(x)) works, rearranging to get sin(x) and using triangles to get cos(x) and then partial fractions, have yet to think of a more elegant method.
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    (Original post by Clarity Incognito)
    A substitution of u=(1-k^2sin^2(x)) works, rearranging to get sin(x) and using triangles to get cos(x) and then partial fractions, have yet to think of a more elegant method.
    Thanks for your message!

    I rearranged the integral to do it in the dxdy direction, I am left with

    sin(2x)*(1-k^2*sin^2(x))^1/2 dx with the limits from 0 to pi/2 but that doesn't help at all.

    The answer is supposed to be

    ((1-k^2)^3/2 -1)/3k^2

    Did you get near that?
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    (Original post by boromir9111)
    Thanks for your message!

    I rearranged the integral to do it in the dxdy direction, I am left with

    sin(2x)*(1-k^2*sin^2(x))^1/2 dx with the limits from 0 to pi/2 but that doesn't help at all.

    The answer is supposed to be

    ((1-k^2)^3/2 -1)/3k^2

    Did you get near that?
    I'm going to concentrate on getting you to the integral of  (1-k^2sin^2(x))^{\frac{1}{2}} and I'll let you do the rest.

     \displaystyle \int (1-k^2sin^2(x))^{\frac{1}{2}} \ dx

     u =1-k^2sin^2(x)

     sin(x) = \dfrac{(1-u)^{\frac{1}{2}}}{k}

    so  cos(x) = \dfrac{(k^2-1+u)^{\frac{1}{2}}}{k}

     \dfrac{du}{dx} = -2k^2sin(x)cos(x)

     \displaystyle -\dfrac{1}{2} \int \dfrac{u^{\frac{1}{2}}}{k^2sin(x  )cos(x)} \ du

     \displaystyle -\dfrac{1}{2} \int \dfrac{u^{\frac{1}{2}}}{\sqrt{(k  ^2-1+u)(1-u)}} \ du

    Partial fractions and then integrate.
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    (Original post by Clarity Incognito)
    I'm going to concentrate on getting you to the integral of  (1-k^2sin^2(x))^{\frac{1}{2}} and I'll let you do the rest.

     \displaystyle \int (1-k^2sin^2(x))^{\frac{1}{2}} \ dx

     u =1-k^2sin^2(x)

     sin(x) = \dfrac{(1-u)^{\frac{1}{2}}}{k}

    so  cos(x) = \dfrac{(k^2-1+u)^{\frac{1}{2}}}{k}

     \dfrac{du}{dx} = -2k^2sin(x)cos(x)

     \displaystyle -\dfrac{1}{2} \int \dfrac{u^{\frac{1}{2}}}{k^2sin(x  )cos(x)} \ du

     \displaystyle -\dfrac{1}{2} \int \dfrac{u^{\frac{1}{2}}}{\sqrt{(k  ^2-1+u)(1-u)}} \ du

    Partial fractions and then integrate.

    Thanks mate. Much appreciated!

    Actually mate, isn't it supposed to be k -1+u and not k^2 -1+u? cos^2(x) = 1-sin^2(x),

    So we have 1 - ((1-u)^1/2)/k etc etc?
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    (Original post by boromir9111)
    Thanks mate. Much appreciated!

    Actually mate, isn't it supposed to be k -1+u and not k^2 -1+u? cos^2(x) = 1-sin^2(x),

    So we have 1 - ((1-u)^1/2)k etc etc?
     cos^2(x)= 1-sin^2x = 1 - \dfrac{1-u}{k^2} = \dfrac{k^2 - 1 + u}{k^2}
 
 
 
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