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Math Help - intergration

  1. #1
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    intergration

    hello!!

    ive got two maths problems and am pretty unsure on both of them so if anyone is bored or wanting to help, it would go a long way

    the first is: evaluate the following intergral, where R is the region of the plane satisfying 0<x2+y2<a

    double intergation sign (1+x/a)dA

    and the second is: evaluate the intergral

    double intergration sign with a infinity sign on the top and a 0 on the bottom with (e^-x^2 dx)

    so the second is like a power to a power thing

    i think that the first is polar coordinates but i am not sure!
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  2. #2
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    Quote Originally Posted by college_clare View Post
    double intergration sign with a infinity sign on the top and a 0 on the bottom with (e^-x^2 dx)
    Same Idea.
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    Quote Originally Posted by ThePerfectHacker View Post

    hmm i dont understand what was happening there. its a bit to complicated to me because the concept of polar coordinates is really new to me. could you explain how i would tackle my problems please?!
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    Quote Originally Posted by college_clare View Post
    hello!!

    ive got two maths problems and am pretty unsure on both of them so if anyone is bored or wanting to help, it would go a long way

    the first is: evaluate the following intergral, where R is the region of the plane satisfying 0<x2+y2<a

    double intergation sign (1+x/a)dA
    Try it this way. We know that x = r cos(\theta) in polar coordinates and the region of integration is the solid disc centered at the origin with r < \sqrt{a}. Thus:
    \int \int _A \left (1 + \frac{x}{a} \right ) dA = \int_0^{\sqrt{a}} \int_0^{2\pi} \left ( 1 + \frac{r cos(\theta)}{a} \right ) r dr d\theta

    I'm sure you can handle the integrations, so I'll just mention:
    \int_0^{\sqrt{a}} \int_0^{2\pi} \left ( 1 + \frac{r cos(\theta)}{a} \right ) r dr d\theta = \int_0^{\sqrt{a}} (2 \pi) r dr = \pi a

    -Dan
    Last edited by topsquark; July 10th 2007 at 06:08 PM.
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    cheers dan, but ive been spending a couple of hours on this one! and i always get stuck on the intergration, could you offer some helping steps to start me off please on the right track?

    and is the pi a2 the final answer?!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by college_clare View Post
    cheers dan, but ive been spending a couple of hours on this one! and i always get stuck on the intergration, could you offer some helping steps to start me off please on the right track?

    and is the pi a2 the final answer?!
    Okay, let's do this again. As Dan said, the integration technique does not change, we only plug in different limits at the end.

    \int_0^{ \sqrt {a}} \int_0^{2\pi} \left ( 1 + \frac{r cos(\theta)}{a} \right ) r dr d\theta

    let's integrate with respect to \theta first. this means we treat everything that does not have a \theta attached to it as a constant, this includes r, for this integration, r is a constant as far as we're concerned.

    thus we get:

    \int_{0}^{ \sqrt {a}} \left[ \theta + \frac {r \sin \theta}{a} \right]_{0}^{2 \pi} ~r~dr

    now plug in the limits, we get:

    \int_{0}^{ \sqrt {a}} 2 \pi r~dr

    so the 2 \pi is a constant, so we forget about it temporarily, integrate r and reattach it. thus we obtain

    \left. \pi r^2 \right|_{0}^{ \sqrt {a}}

    again, using the fundamental theorem of calculus to evaluate between the limits, we end up with the final answer

    \pi a

    ...technically it should be \pi |a|, but since a>0 there is no harm in writing \pi a
    Last edited by Jhevon; July 11th 2007 at 05:06 AM. Reason: Making the correction Dan suggested to the limits of integration. Thanks Dan
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    Forum Admin topsquark's Avatar
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    I'm afraid there was a slight error in my work and it threw Jhevon off, too.

    The region of integration is 0 < x^2 + y^2 < a which is a solid disc centered on the origin of radius \sqrt{a}, not "a" like I had said in my previous post. This means the result of the integration is \pi a, not \pi a^2. (I have fixed this in my original post.)

    Jhevon's solution is perfectly valid once this same change is made and this change does not affect the integration technique.

    -Dan
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  8. #8
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    Jhevon and Topsquart, i am so sorry but i really dont understand what is happening as polar coordinates is really new to me. how did you get the new limits initially and how did you go about the intergration? sorry i dont understnad
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by college_clare View Post
    Jhevon and Topsquart, i am so sorry but i really dont understand what is happening as polar coordinates is really new to me. how did you get the new limits initially and how did you go about the intergration? sorry i dont understnad
    Take a look at the region of integration:
    0 < x^2 + y^2 < a

    Since the equation of a circle of radius \sqrt{a} centered on the origin is x^2 + y^2 = a we know that the region of integration is a solid disc of radius \sqrt{a} centered on the origin.

    How do you cover that area in polar coordinates? Well obviously the limits on the "radius" r are going to be 0 and \sqrt{a}. Since the entire region is covered (ie. we don't have a slice of a circle or anything) we know the limits on the angular variable \theta are 0 and 2\pi.

    As far as how the dA becomes r dr d\theta I'm going to leave that one to your Calculus text as I can't draw the diagram and do it justice.

    Jhevon did a good job explaining how to integrate this. I'll go over the \theta integration again, but if you are having significant difficulties do these then you need to brush up on your integration techniques.

    \int_0^{2\pi} \left ( 1 + \frac{r cos(\theta)}{a} \right ) r d\theta

    I am going to integrate over \theta first. Why? Because I know it's going to give me an easier time than doing the r integration first. Seeing that is just a matter of experience and you can certainly do the problem with the r integration first.

    When we do this we are considering r to be a constant. So how would you do this integral?
    \int_0^{2\pi} \left ( 1 + \frac{r cos(\theta)}{a} \right ) r d\theta

    = \int_0^{2\pi} r d\theta + \int_0^{2\pi} \frac{r^2 cos(\theta)}{a} d\theta

    = r \int_0^{2\pi} d\theta + \frac{r^2}{a} \int_0^{2\pi} cos(\theta) d\theta
    (Remember that r is a constant for now.)

    Both integrations are trivial:
    = r \theta |_0^{2\pi} + \frac{r^2}{a}  \cdot sin(\theta)|_0^{2\pi}

    = r (2\pi - 0) + \frac{r^2}{a}  \cdot (sin(2\pi) - sin(0))

    = 2\pi r + \frac{r^2}{a}  \cdot (0 - 0)

    = 2\pi r

    This then is the integrand for the r integration, which I'm sure you can do.

    -Dan
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