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Help with Change of Variable in Multiple Integral Question watch

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    An ellipse is given by

    \frac{x^2}{a} +\frac{y^2}{b} = 1

    You want to find the area by using a change of coordinates:
    x = r cos θ, y =\frac{br}{a}sin θ.
    Find the range of values of r and θ that correspond to the interior of the ellipse.

    Find the Jacobian of the transformation.
    Find the area.

    Now the Jacobian I got was J = \frac{br}{a}

    Now the problem is..... I don't know what to integrate...is it:

    \displaystyle\int^\pi_0\int^1_0 x^{2} + y^{2}\ dx dy

    then substituting in x = r cos θ, y =\frac{br}{a}sin θ

    \displaystyle\int^\pi_0 \int^1_0 [(rcos\theta)^{2} + (\frac{br}{a}sin^{2}\theta)]\frac{br}{a}drd\theta

    Because I haven't been given the function so I assumed it was integrating
    x^{2} + y^{2}...Please check if my limits are right they are (2pi and 0) for the first integral....Don't know how to write 2pi in the integral using latex yet!

    And if this is right....why can't I get the correct answer?

    Thanks
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    (Original post by mathsRus)
    An ellipse is given by

    \frac{x^2}{a} +\frac{y^2}{b} = 1

    You want to find the area by using a change of coordinates:
    x = r cos θ, y =\frac{br}{a}sin θ.
    Find the range of values of r and θ that correspond to the interior of the ellipse.

    Find the Jacobian of the transformation.
    Find the area.

    Now the Jacobian I got was J = \frac{br}{a}

    Now the problem is..... I don't know what to integrate...is it:

    \displaystyle\int^\pi_0\int^1_0 x^{2} + y^{2}\ dx dy

    then substituting in x = r cos θ, y =\frac{br}{a}sin θ

    \displaystyle\int^\pi_0 \int^1_0 [(rcos\theta)^{2} + (\frac{br}{a}sin^{2}\theta)]\frac{br}{a}drd\theta

    Because I haven't been given the function so I assumed it was integrating
    x^{2} + y^{2}...Please check if my limits are right they are (2pi and 0) for the first integral....Don't know how to write 2pi in the integral using latex yet!

    And if this is right....why can't I get the correct answer?

    Thanks
    post a photo of the question (saves you type too)
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    (Original post by TeeEm)
    post a photo of the question (saves you type too)
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    here it is
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    (Original post by mathsRus)
    Name:  qu.png
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Size:  90.7 KB

    here it is
    firstly apologies I refuse to use LaTeX

    the area is the Integral of 1 dA, where dA is the area element

    in other words dA = J dr dθ

    the transformation they give you makes the ellipse into a circle of radius a

    to sweep this area you need

    r=0 to r=a

    and

    θ=0 to θ=2π
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    (Original post by TeeEm)
    firstly apologies I refuse to use LaTeX

    the area is the Integral of 1 dA, where dA is the area element

    in other words dA = J dr dθ

    the transformation they give you makes the ellipse into a circle of radius a

    to sweep this area you need

    r=0 to r=a

    and

    θ=0 to θ=2π

    So is it?
    A = \displaystyle\int^{2\pi} _0 \int^a_0\frac{br}{a}drd\theta
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    (Original post by mathsRus)
    So is it?
    A = \displaystyle\int^{2\pi} _0 \int^1_0\frac{br}{a}drd\theta

    almost

    r goes from 0 to a
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    (Original post by TeeEm)

    almost

    r goes from 0 to a
    yup still getting used to \mathrm{Latex!}

    I still don't understand how this question kind of transforms it into a circle
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    (Original post by mathsRus)
    yup still getting used to \mathrm{Latex!}

    I still don't understand how this question kind of transforms it into a circle
    I was hoping you would not ask ....

    well

    first multiply the ellipse as given by a2

    x2 + (a2y2)/b2

    then X = x, Y =( ay/b )

    X2 + Y2 = a2
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    (Original post by TeeEm)
    I was hoping you would not ask ....

    well

    first multiply the ellipse as given by a2

    x2 + (a2y2)/b2

    then X = x, Y =( ay/b )

    X2 + Y2 = a2
    wow that is some trick of playing around I will have to learn haha

    Thanks
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    (Original post by mathsRus)
    wow that is some trick of playing around I will have to learn haha

    Thanks
    no worries...
    keep practicing
 
 
 
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