find volume using Cavalieri - difficult integral












1












$begingroup$


$B_n(0,1)$ is the $n$ dimensional ball, centered at zero with radius one.



We define the set $P_n = {x in B_n: |x_1| < frac{1}{1000}}$. Find $v(P_n)$.



What I did:



Notice that the intersection of $P_n$ with the plane $x_1 = t$ is well defined for $t in (-frac{1}{1000}, frac{1}{1000})$ and that $P_n cap{x in mathbb R^n: x = t} = {x in mathbb R^n: x_2^2+x_3^2+ dots +x_n^2 < 1-t^2} =B_{n-1}(0,sqrt{1-t^2})$.



Now from Cavalieri's principle:



$$v(P_n) = int_{P_n}1dx = int_{-frac{1}{1000}}^{frac{1}{1000}}int_{B_{n-1}(0,sqrt{1-t^2})}1dxdt = v(B_{n-1}(0,1))int_{-frac{1}{1000}}^{frac{1}{1000}}(1-t^2)^{frac{n-1}{2}}dt$$



How would we calculate this integral?






calculus integration definite-integrals







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$endgroup$








  • 3




    $begingroup$
    Substitute t = $sin(theta)$ into the integral, and use a identity for the integral of $cos(theta)^n$.
    $endgroup$
    – user458276
    Dec 29 '18 at 21:37
















1












$begingroup$


$B_n(0,1)$ is the $n$ dimensional ball, centered at zero with radius one.



We define the set $P_n = {x in B_n: |x_1| < frac{1}{1000}}$. Find $v(P_n)$.



What I did:



Notice that the intersection of $P_n$ with the plane $x_1 = t$ is well defined for $t in (-frac{1}{1000}, frac{1}{1000})$ and that $P_n cap{x in mathbb R^n: x = t} = {x in mathbb R^n: x_2^2+x_3^2+ dots +x_n^2 < 1-t^2} =B_{n-1}(0,sqrt{1-t^2})$.



Now from Cavalieri's principle:



$$v(P_n) = int_{P_n}1dx = int_{-frac{1}{1000}}^{frac{1}{1000}}int_{B_{n-1}(0,sqrt{1-t^2})}1dxdt = v(B_{n-1}(0,1))int_{-frac{1}{1000}}^{frac{1}{1000}}(1-t^2)^{frac{n-1}{2}}dt$$



How would we calculate this integral?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Substitute t = $sin(theta)$ into the integral, and use a identity for the integral of $cos(theta)^n$.
    $endgroup$
    – user458276
    Dec 29 '18 at 21:37














1












1








1





$begingroup$


$B_n(0,1)$ is the $n$ dimensional ball, centered at zero with radius one.



We define the set $P_n = {x in B_n: |x_1| < frac{1}{1000}}$. Find $v(P_n)$.



What I did:



Notice that the intersection of $P_n$ with the plane $x_1 = t$ is well defined for $t in (-frac{1}{1000}, frac{1}{1000})$ and that $P_n cap{x in mathbb R^n: x = t} = {x in mathbb R^n: x_2^2+x_3^2+ dots +x_n^2 < 1-t^2} =B_{n-1}(0,sqrt{1-t^2})$.



Now from Cavalieri's principle:



$$v(P_n) = int_{P_n}1dx = int_{-frac{1}{1000}}^{frac{1}{1000}}int_{B_{n-1}(0,sqrt{1-t^2})}1dxdt = v(B_{n-1}(0,1))int_{-frac{1}{1000}}^{frac{1}{1000}}(1-t^2)^{frac{n-1}{2}}dt$$



How would we calculate this integral?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$




$B_n(0,1)$ is the $n$ dimensional ball, centered at zero with radius one.



We define the set $P_n = {x in B_n: |x_1| < frac{1}{1000}}$. Find $v(P_n)$.



What I did:



Notice that the intersection of $P_n$ with the plane $x_1 = t$ is well defined for $t in (-frac{1}{1000}, frac{1}{1000})$ and that $P_n cap{x in mathbb R^n: x = t} = {x in mathbb R^n: x_2^2+x_3^2+ dots +x_n^2 < 1-t^2} =B_{n-1}(0,sqrt{1-t^2})$.



Now from Cavalieri's principle:



$$v(P_n) = int_{P_n}1dx = int_{-frac{1}{1000}}^{frac{1}{1000}}int_{B_{n-1}(0,sqrt{1-t^2})}1dxdt = v(B_{n-1}(0,1))int_{-frac{1}{1000}}^{frac{1}{1000}}(1-t^2)^{frac{n-1}{2}}dt$$



How would we calculate this integral?






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 29 '18 at 21:39









Bernard

119k639112




119k639112










asked Dec 29 '18 at 21:32









Oria GruberOria Gruber

6,50732360




6,50732360








  • 3




    $begingroup$
    Substitute t = $sin(theta)$ into the integral, and use a identity for the integral of $cos(theta)^n$.
    $endgroup$
    – user458276
    Dec 29 '18 at 21:37














  • 3




    $begingroup$
    Substitute t = $sin(theta)$ into the integral, and use a identity for the integral of $cos(theta)^n$.
    $endgroup$
    – user458276
    Dec 29 '18 at 21:37








3




3




$begingroup$
Substitute t = $sin(theta)$ into the integral, and use a identity for the integral of $cos(theta)^n$.
$endgroup$
– user458276
Dec 29 '18 at 21:37




$begingroup$
Substitute t = $sin(theta)$ into the integral, and use a identity for the integral of $cos(theta)^n$.
$endgroup$
– user458276
Dec 29 '18 at 21:37










1 Answer
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$begingroup$

I'll elaborate on @user458276's comment: For an integral $int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t$ first substitute $t = sintheta$ to obtain



$$int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t = int_{-arcsin a}^{arcsin a} (cos^2 theta)^frac{n-1}{2} cos theta ,mathrm{d} theta = 2 int_0^{arcsin a} cos^n theta , mathrm{d} theta$$



Then take a look at a list of trig integrals and find



$$ int cos^n theta ,mathrm{d}theta = frac1{n} cos^{n-1}theta ,sintheta - frac{n-1}{n}int cos^{n-2} theta ,mathrm{d}theta$$



Since $n$ is an integer repeatedly applying this expression will lead to a known integral ($int mathrm{d}theta$ for even $n$ and $intcostheta,mathrm{d}theta$ for odd $n$).






share|cite|improve this answer











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    1 Answer
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    $begingroup$

    I'll elaborate on @user458276's comment: For an integral $int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t$ first substitute $t = sintheta$ to obtain



    $$int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t = int_{-arcsin a}^{arcsin a} (cos^2 theta)^frac{n-1}{2} cos theta ,mathrm{d} theta = 2 int_0^{arcsin a} cos^n theta , mathrm{d} theta$$



    Then take a look at a list of trig integrals and find



    $$ int cos^n theta ,mathrm{d}theta = frac1{n} cos^{n-1}theta ,sintheta - frac{n-1}{n}int cos^{n-2} theta ,mathrm{d}theta$$



    Since $n$ is an integer repeatedly applying this expression will lead to a known integral ($int mathrm{d}theta$ for even $n$ and $intcostheta,mathrm{d}theta$ for odd $n$).






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      I'll elaborate on @user458276's comment: For an integral $int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t$ first substitute $t = sintheta$ to obtain



      $$int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t = int_{-arcsin a}^{arcsin a} (cos^2 theta)^frac{n-1}{2} cos theta ,mathrm{d} theta = 2 int_0^{arcsin a} cos^n theta , mathrm{d} theta$$



      Then take a look at a list of trig integrals and find



      $$ int cos^n theta ,mathrm{d}theta = frac1{n} cos^{n-1}theta ,sintheta - frac{n-1}{n}int cos^{n-2} theta ,mathrm{d}theta$$



      Since $n$ is an integer repeatedly applying this expression will lead to a known integral ($int mathrm{d}theta$ for even $n$ and $intcostheta,mathrm{d}theta$ for odd $n$).






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        I'll elaborate on @user458276's comment: For an integral $int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t$ first substitute $t = sintheta$ to obtain



        $$int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t = int_{-arcsin a}^{arcsin a} (cos^2 theta)^frac{n-1}{2} cos theta ,mathrm{d} theta = 2 int_0^{arcsin a} cos^n theta , mathrm{d} theta$$



        Then take a look at a list of trig integrals and find



        $$ int cos^n theta ,mathrm{d}theta = frac1{n} cos^{n-1}theta ,sintheta - frac{n-1}{n}int cos^{n-2} theta ,mathrm{d}theta$$



        Since $n$ is an integer repeatedly applying this expression will lead to a known integral ($int mathrm{d}theta$ for even $n$ and $intcostheta,mathrm{d}theta$ for odd $n$).






        share|cite|improve this answer











        $endgroup$



        I'll elaborate on @user458276's comment: For an integral $int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t$ first substitute $t = sintheta$ to obtain



        $$int_{-a}^a (1 - t^2)^{frac{n-1}{2}} mathrm{d}t = int_{-arcsin a}^{arcsin a} (cos^2 theta)^frac{n-1}{2} cos theta ,mathrm{d} theta = 2 int_0^{arcsin a} cos^n theta , mathrm{d} theta$$



        Then take a look at a list of trig integrals and find



        $$ int cos^n theta ,mathrm{d}theta = frac1{n} cos^{n-1}theta ,sintheta - frac{n-1}{n}int cos^{n-2} theta ,mathrm{d}theta$$



        Since $n$ is an integer repeatedly applying this expression will lead to a known integral ($int mathrm{d}theta$ for even $n$ and $intcostheta,mathrm{d}theta$ for odd $n$).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 29 '18 at 22:04

























        answered Dec 29 '18 at 21:55









        0x5390x539

        1,067317




        1,067317






























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