Results 1 to 13 of 13

Math Help - Integral Area

  1. #1
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Integral Area

    I don't really understand this question:
    Evaluate:
    \int_0^3 x \sqrt{81-x^4}\, dx

    by making a substitution of the form u=x^p and then interpreting the result in terms of a known area.

    I get that I should find a usable u=x^p then evaluate but don't understand what it means by interpreting in terms of a known area.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,698
    Thanks
    454

    Re: Integral Area

    Quote Originally Posted by Bowlbase View Post
    I don't really understand this question:
    Evaluate:
    \int_0^3 x \sqrt{81-x^4}\, dx

    by making a substitution of the form u=x^p and then interpreting the result in terms of a known area.

    I get that I should find a usable u=x^p then evaluate but don't understand what it means by interpreting in terms of a known area.

    Thanks
    u = x^2

    du = 2x \, dx

    \frac{1}{2} \int_0^3 2x \sqrt{81-x^4} \, dx

    substitute and reset the limits of integration ...

    \frac{1}{2} \int_0^9 \sqrt{81-u^2} \, du

    now, what common geometric figure is formed by the graph of y = \sqrt{81-u^2} ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Re: Integral Area

    semi-circle. So, this would the area y=0 to y=9?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425

    Re: Integral Area

    Quote Originally Posted by Bowlbase View Post
    semi-circle. So, this would the area y=0 to y=9?
    Surely it would be the area of a quarter of the circle 9 units in radius centred at the origin...

    If you want to use integrals, then make the substitution \displaystyle u = \sin{\theta} \implies du = \cos{\theta}\,d\theta...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,698
    Thanks
    454

    Re: Integral Area

    Quote Originally Posted by Prove It View Post
    Surely it would be the area of a quarter of the circle 9 units in radius centred at the origin...
    don't forget the \frac{1}{2} in front of the definite integral ...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Re: Integral Area

    Okay, I think I understand. If I take 1/8 of pir^2 I should have the correct answer then, no?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425

    Re: Integral Area

    Quote Originally Posted by Bowlbase View Post
    Okay, I think I understand. If I take 1/8 of pir^2 I should have the correct answer then, no?
    Yes. You might also want to try solving the integral directly as well, just to verify your answer
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Re: Integral Area

    Yes, of course. I doubt my instructor would accept it otherwise.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Re: Integral Area

    Okay, I thought I had this sorted out last night but obviously didn't think too much as I worked this problem.

    First:
    Having gotten \frac{1}{2} \int_0^9 \sqrt{81-u^2} \, du
    It's not clear to me why we're changing the range from 3 to 9. From a geometric point of view it makes perfect sense but just taking it in the context of the integral I don't see the reasoning.

    Second: I thought that it would be simple to evaluate this at first but we have not done anything with squares inside roots. Do I need to make another substitution for this? For instance c=u^2?

    Thanks for the help so far!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425

    Re: Integral Area

    When you change the integral from an integral with respect to x to an integral with respect to u, the terminals need to change to u values.

    Since \displaystyle \begin{align*} u = x^2  \end{align*} , when \displaystyle \begin{align*} x = 0, u = 0  \end{align*} and when \displaystyle \begin{align*} x = 3, u = 9  \end{align*} .

    Now, as for solving the integral, like I suggested, you need to make the substitution \displaystyle \begin{align*}   u = 9\sin{\theta} \implies du = 9\cos{\theta}\,d\theta \end{align*} , and note that when \displaystyle \begin{align*}   u = 0, \theta = 0\end{align*} and when \displaystyle \begin{align*} u = 9, \theta = \frac{\pi}{2}  \end{align*} and the integral becomes...

    \displaystyle \begin{align*} \frac{1}{2}\int_0^9{ \sqrt{ 81 - u^2 } \,du } &= \frac{1}{2}\int_0^{ \frac{\pi}{2} }{ \sqrt{ 81 - \left( 9\sin{\theta} \right)^2 }\,9\cos{\theta} \, d\theta } \\ &= \frac{9}{2}\int_0^{\frac{\pi}{2}} { \sqrt{ 81 - 81 \sin^2{\theta} }  \,\cos{\theta}\,d\theta} \\ &= \frac{9}{2}\int_0^{\frac{\pi}{2}} { \sqrt{ 81\left(1 - \sin^2{\theta}\right)  } \,\cos{\theta}\,d\theta} \\ &= \frac{9}{2}\int_0^{\frac{\pi}{2}} { \sqrt{ 81\cos^2{\theta} } \,\cos{\theta}\,d\theta} \\ &= \frac{9}{2}\int_0^{\frac{\pi}{2}}{ 9\cos{\theta}\cos{\theta}\,d\theta} \\ &= \frac{81}{2}\int_0^{\frac{\pi}{2}} {\cos^2{\theta}\,d\theta} \\ &= \frac{81}{2}\int_0^{\frac{\pi}{2}}{\frac{1}{2} + \frac{1}{2}\cos{2\theta}\,d\theta} \\ &= \frac{81}{2}\left[\frac{1}{2}\theta + \frac{1}{4}\sin{2\theta} \right]_0^{\frac{\pi}{2}} \end{align*}

    \displaystyle \begin{align*} &= \frac{81}{2}\left[\left(\frac{1}{2}\cdot \frac{\pi}{2} + \frac{1}{4}\sin{2\cdot \frac{\pi}{2}}\right) - \left(\frac{1}{2}\cdot 0 + \frac{1}{4}\sin{2\cdot 0}\right)\right] \\ &= \frac{81}{2} \left[\left( \frac{\pi}{4} + \frac{1}{4}\sin{\pi}\right) - \left(0 + \frac{1}{4}\sin{0}\right)\right] \\ &= \frac{81\pi}{8}   \end{align*}

    which is the same as half of the area of a quarter circle of radius 9 as discussed.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Re: Integral Area

    We have yet to introduce a trig function into a integral, or derivative for that matter, that did not already have one in my class. So I never would have even guessed to do this...

    Are there rules that precede doing this? Are we able to do this because it is part of a circular function?
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425

    Re: Integral Area

    Quote Originally Posted by Bowlbase View Post
    We have yet to introduce a trig function into a integral, or derivative for that matter, that did not already have one in my class. So I never would have even guessed to do this...

    Are there rules that precede doing this? Are we able to do this because it is part of a circular function?
    I suggest reading this to learn about Trigonometric Substitution.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member
    Joined
    Jan 2011
    Posts
    52

    Re: Integral Area

    Quote Originally Posted by Prove It View Post
    I suggest reading this to learn about Trigonometric Substitution.

    This helped out a lot, thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. area integral on a circle
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 1st 2012, 05:39 PM
  2. Integral Area
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 4th 2011, 07:06 PM
  3. Area Integral
    Posted in the Calculus Forum
    Replies: 13
    Last Post: January 28th 2011, 04:37 AM
  4. How to write area as integral?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 17th 2009, 09:23 PM
  5. Area Integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 19th 2008, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum