Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trigonometric Substitution- Area help

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Let R be the smaller of the two regions enclosed by the elipse 144 x2+64 y2=9216 and the line [tex]$ x=(8 \sqrt{2})/2$[/tex].
    Find the area of the region R.

    2. Relevant equations

    3. The attempt at a solution
    my textbook doesnt have anything like this.. i have no idea where to start even.
    please help!
  2. jcsd
  3. Feb 27, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    Sketch the curve and the line. This is really just a plain 'area under the curve problem'. There's nothing exotic about it.
  4. Feb 27, 2010 #3
    her is the image of the graph:
    http://img521.imageshack.us/img521/5048/ellipse.png [Broken]

    so xR = 4*[tex]$\sqrt{2}$[/tex]
    xL = [tex]$\sqrt{(576-4y^2)/9}$[/tex] (i reduced down all the numbers of the elipse by 16)

    and now just find A(x) = [tex]$\int_{b'}^b (XR - XL)dx$[/tex]

    Last edited by a moderator: May 4, 2017
  5. Feb 27, 2010 #4


    User Avatar
    Science Advisor
    Homework Helper

    Well, ok. What you are describing is how you would do it by integrating dy, not dx. But how can you 'reduce numbers by 16'?? Don't do that. x=sqrt(9216-64*y^2). Now you just need to find b and b' and do the integration. You might find it a little easier to integrate dx instead. Solve for y in terms of x and integrate from x=4*sqrt(2) to the end of the ellipse.
  6. Feb 28, 2010 #5
    well in other words, i did everything right except the 'reducing by 16' part.. right?
    all i gotta do now is plug my x's into the
    A= [tex] $\int_{b'}^b (XR - XL)dx$ [/tex]

    find the antiderivative and im good to go... correct me if im wrong please!
  7. Feb 28, 2010 #6


    User Avatar
    Science Advisor
    Homework Helper

    Right, except doing it the way you've outlined means you are integrating dy. You need to find y bounds.
  8. Feb 28, 2010 #7
    ok im lost now.
    there is this equation for elipse (x^2 / a^2) + (y^2 / b^2) = 1
    so i get (x^2 / 64) + (y^2 / 144) = 1
    then i solve for y and i get
    y = +- sqrt(144- (9x^2)/4 )
    so the y+ is the upper bound and y- is the lower bound.

    apparently im supposed to use the substitution 'asint' well because its Trigonometric Substitution lol

    what confused me is that you said im right but then the students from my class say other things on the online assignment website.

    A = [tex] $\int_{-\sqrt(144- (9x^2)/4 )}^{\sqrt(144- (9x^2)/4 )} (\sqrt(9216-64*y^2) - 4*\sqrt{2})dy$ [/tex]

    or am i supposed to do yT - yB? cuz from the graph it looks like it should be the X's not Y's.. if you know what i mean by that
    Last edited: Feb 28, 2010
  9. Feb 28, 2010 #8


    User Avatar
    Science Advisor
    Homework Helper

    You can do it either dx or dy. It will work either way. Your problems now are i) you are STILL writing dx instead of dy and ii) your upper and lower limits should be numbers. They are the y coordinates of the points where the line and the ellipse intersect.
  10. Feb 28, 2010 #9
    oh right i forgot to change it to dy..
    i was thinking doing this:
    2*A = [tex] $\int_{0}^{\sqrt(144- (9x^2)/4 )} (\sqrt(9216-64*y^2) - 5.65685)dy$ [/tex]

    how do i get a # for the upper bound? i mean what x-value do i plug in there?
    im happy that im on the right truck at least lol
  11. Feb 28, 2010 #10


    User Avatar
    Science Advisor
    Homework Helper

    To find the bounds you intersect x=4*sqrt(2) with the ellipse. Just substitute x=4*sqrt(2) into the ellipse equation and find the corresponding y.
    Last edited: Feb 28, 2010
  12. Feb 28, 2010 #11
    oh okay.. so i get
    2A = [tex] $\int_{0}^{8.48528} (\sqrt(9216-64y^2)dy - 48 + C$[/tex]

    y = 96sint
    dy = 96cost dt

    2A = [tex] $\int_{0}^{8.48528} [\sqrt(9216-64(96sint)^2]*96cost dt - 48 + C$[/tex]

    something is wrong.. i feel like im missing a step or something, can you tell me what am i missing?
  13. Feb 28, 2010 #12


    User Avatar
    Science Advisor
    Homework Helper

    Something is wrong. Why do you think 96*sin(t) is a good substitution? And can you show more of your steps? I had a hard time figuring out where the 48 came from. Especially since you are changing limits to decimal. And there's no call for a '+C'. These are definite integrals.
  14. Feb 28, 2010 #13
    sorry about that.. i just got excited when i got the answer for that part so i just put the number in.
    [tex]$\sqrt(9216-64y^2)$[/tex] = [tex]$\sqrt(96^2-(8y)^2)$[/tex]

    and as that substitution rule states: [tex]$\sqrt(a^2-x^2)$[/tex]
    => x= asint
    dx = acostdt
    so [tex]$\sqrt(a^2-(asint)^2)$[/tex]

    y= 96sint
    dy = 96costdt

    which leads to
    [tex] $\int_{0}^{8.48528} [\sqrt(9216-(8*96sint)^2]*96cost dt - 48$ [/tex]
  15. Feb 28, 2010 #14


    User Avatar
    Science Advisor
    Homework Helper

    The 'x' in your subsitution rule is 8y. You want 8y=96*sin(t).
  16. Feb 28, 2010 #15
    ok so i just found a mistake in my XR calculation.
    it's actually
    [tex] $\sqrt((9216-64y^2)/144)$ [/tex] = [tex] $\sqrt(64-(4y^2)/9)$ [/tex] =


    cuz i forgot to divide the whole thing by 144.

    so now, 2y/3 = 8sint
    dy = 12costdt

    A= [tex]
    $\int_{-8.48528}^{8.48528} ([\sqrt(64-(8sint)^2 ] - 4\sqrt(2) ) dy$

    which leads to
    [tex] $\int_{-8.48528}^{8.48528} (8cost - 4\sqrt(2) ) dy$ [/tex]
    =[tex] $\int_{-8.48528}^{8.48528} (8cost - 4\sqrt(2) ) 12costdt$ [/tex]
    =12 [tex] $\int_{-8.48528}^{8.48528} (8cos^2t - 4\sqrt(2) cost)dt$ [/tex]
    =12*4 [tex] $\int_{-8.48528}^{8.48528} (2cos^2t - \sqrt(2) cost)dt$ [/tex]
    =48*( [tex] $ 2 *\int_{-y}^{y} (cos^2t)dt - \sqrt(2)*\int_{-y}^{y} ( cost)dt$ [/tex])

    =48*( [tex] $ 2 *\int_{-y}^{y} (0.5(1-cos2t))dt - \sqrt(2)*sint$ [/tex])

    =48*( [tex] $ \int_{-y}^{y} (1-cos2t)dt - \sqrt(2)*sint$ [/tex])

    is anything wrong with it?
    i checked it over many times already..
    my question is now what do i do once i find the antiderivatives.. i mean, change sin/cos with whatever i substituted it to?
    Last edited: Feb 28, 2010
  17. Feb 28, 2010 #16


    User Avatar
    Science Advisor
    Homework Helper

    You are really losing me here. How did the trig functions get mixed up the the 4*sqrt(2) part? I thought you already decided that was 48? Why don't you back up and make a fresh start and present all of your steps in order. I really can't follow that. I just can't. You are doing some stuff right, but the organization is just so messed up.
  18. Feb 28, 2010 #17
    xR = [tex]$\sqrt((9216-64y^2)/144)$[/tex]
    xL = 4 [tex] $\sqrt{2}$ [/tex] = 5.65685

    i got the bounds a = -8.485, b = 8.485

    A=[tex] $\int_{a}^{b} (xr-xl)dy $[/tex]

    A=[tex] $\int_{a}^{b} ((\sqrt((9216-64y^2)/144))-5.65685)dy $[/tex]

    A=[tex] $\int_{a}^{b} ((\sqrt((64-(4/9)y^2)))-5.65685)dy $[/tex]

    A=[tex] $\int_{a}^{b} ((\sqrt((8^2-(2y/3)^2)))-5.65685)dy $[/tex]

    A=[tex] $\int_{a}^{b} \sqrt(8^2-(2y/3)^2)dy - $\int_{a}^{b}5.65685dy $[/tex]

    (2y/3) = 8sint

    A=[tex] $\int_{a}^{b} \sqrt(64-(8sint)^2)dy - 5.65685y|_{a}^{b} $[/tex]

    A=[tex] $\int_{a}^{b} \sqrt(64-64sin^2t dy - 5.65685y|_{a}^{b} $[/tex]

    A=[tex] $\int_{a}^{b} \sqrt(64(1-sin^2t)) dy - 5.65685y|_{a}^{b} $[/tex]

    A=[tex] $\int_{a}^{b} \sqrt(64(cos^2t)) dy - 5.65685y|_{a}^{b} $[/tex]

    A=[tex] $\int_{a}^{b} 8costdy - 5.65685y|_{a}^{b} $[/tex]

    A=[tex] $ 8sint|_{a}^{b} - 5.65685y|_{a}^{b} $[/tex]

    recall that (2y/3) = 8sint
    A=[tex] $ 2y/3 |_{a}^{b} - 5.65685y|_{a}^{b} $[/tex]

    and i get A=84.6862 but its wrong!

    if i do 0.5A=[tex] $ 2y/3 |_{0}^{8.485} - 5.65685y|_{0}^{8.485} $[/tex] i still get 84.6862
    Last edited: Feb 28, 2010
  19. Feb 28, 2010 #18


    User Avatar
    Science Advisor
    Homework Helper

    Now you dropped the cos(t) from the dy=12costdt, and you've gone back to changing xR from sqrt(9216-64y^2) to sqrt((9216-64y^2)/144). They AREN'T the same thing. You are changing things you corrected back into the old incorrect things. If you are as tired as I am right now I think you should get some rest and try to sort things out in the morning.
  20. Feb 28, 2010 #19
    i dont have til the morning.. its due 8am :'(
    how did you get sqrt(9216-64y^2) in the first place?

    [tex]$\int_{a}^{b} \sqrt(8^2-(2y/3)^2)*12costdt- $ (5.65685)*\int_{a}^{b} 12costdt $[/tex]

    [tex]$ 12*\int_{a}^{b} \sqrt(64 - 4y^2/9)costdt - $ (67.88)sint |_a^b $[/tex]

    im really sorry for the headache i give you. our teacher just expects us to know all this without teaching us so what can i do..
  21. Feb 28, 2010 #20


    User Avatar
    Science Advisor
    Homework Helper

    Oops, you got me. Sure xR=sqrt(64-(2y/3)^2). The problem is I keep flipping back to your original post and the 144 is on the first line and I keep forgetting it's there. Sorry. But I'm still having a hard time tracking here. I am integrating sqrt(64-(2y/3)^2)-4*sqrt(2))*dy from -6*sqrt(2) to +6*sqrt(2) and I'm not getting 84.6862. That's about all I've got in me right now.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook