Matlab s Numerical Integration Commands
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- Marvin Bryan
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1 Mtlb s Numericl Integrtion Commnds The relevnt commnds we consider re qud nd dblqud, triplequd. See the Mtlb help files for other integrtion commnds. By the wy, qud refers to dptive qudrture. To integrte: b Q=qud(function,, b); Q=qud(function,, b, tol); f(x) dx, we use the commnd: The second version of the qud function is used if there is user-defined tolernce. The defult tolernce is 1 6. The only slightly tricky thing here is in how to define nd pss in the function f(x). We hve severl options, deping on the complexity of your definition of f(x). Below we give three exmples: Exmple: which in Mtlb would be: 2 f1=inline( 1./(2*x.^3-2*x-5) ); Q=qud(f1,,2); 1 2x 3 2x 5 dx We could lso define the function using nonymous hndles (See Mtlb documenttion for more): f2=@(x)1./(2*x.^3-2*x-5); Q=qud(f2,,2); We could lso define the function using n M-file. Here is smple, which we ll sve s myfun.m: function y=myfun(x) y=1./(2*x.^3-2*x-5); In the Mtlb commnd window, we would type: Q=qud(@myfun,,2); 1
2 Similrly, if you re integrting using built-in function like: 5 sin(x) dx you would type: Q=qud(@sin,, 5); Suppose we wish to integrte: Double Integrls x 2 y dx dy As before, we cn use either n inline function, n nonymous function hndle, or n M-file: Using n inline function. When defining function of more thn one vrible, we cn list the order in which they should pper t the : f1=inline( x.^2.*y, x, y ); Q=dblqud(f,,3,1,2); Using n nonymous function hndle: f2=@(x,y)x.^2.*y; Q=dblqud(f2,,3,1,2); Using n M-file. First type nd sve the M-file (here, we cll it myfun.m gin): function z=myfun(x,y) z=x.^2.*y; Then cll it in Mtlb Q=dblqud(@myfun,,3,1,2); The double qudrture formul only works on rectngulr regions in the plne. How would we integrte over non-squre regions? It deps on the complexity of the region. Here re two options- The first is where the region is somewht simple ( circle in the plne), nd the second is more generl. 2
3 1. The volume of hemisphere: 1 x 2 1 x 2 1 (x 2 + y 2 ) dy dx In this cse, the integrnd is somewht simple, nd we cn ext the definition of the integrnd so tht it is zero outside the re of interest. Mthemticlly speking, where 1 x 2 1 x 2 G(x, y) = 1 (x 2 + y 2 ) dy dx = G(x, y) dy dx { 1 (x 2 + y 2 ) if x 2 + y 2 1 otherwise In Mtlb, the expression (x.^2+y.^2<=1) is logicl sttement- It returns 1 if the expression is TRUE, if FALSE. Therefore, to find the volume, we could type: f3=@(x,y)sqrt(1-(x.^2+y.^2)).*(x.^2+y.^2<=1); Q=dblqud(f3,-1,1,-1,1); 2. Alterntively (or more generlly), given: b y=g2 (x) ( b ) g2 (x) F (x, y) dy dx = F (x, y) dy dx = y=g 1 (x) g 1 (x) b G(x) dx Tht is, given numericl vlue of x, we would write the function G(x) s: Here s specific exmple: G(x) = g2 (x) g 1 (x) 2 2x x 2 F (x, y) dy x 2 + y 2 dy dx We write the M-file for the inside integrl: function z=myfun(x) n=length(x); z=zeros(size(x)); for j=1:n =x(j)^2; b=2*x(j); h=@(y)(+y.^2); z(j)=qud(h,,b); 3
4 In Mtlb, type: nd you my get some wrnings, but we do get n nswer (the exct vlue is 216/33). 3. Here s nother exmple of this second type, where we build function to evlute the inside integrl: Evlute the volume of the solid under the surfce z = x 3 y 4 + xy 2 nd bove the region bounded by the curves y = x 3 x nd y = x 2 + x for x. Doing our norml clculus thing, we see tht the region of integrtion is for x rnging from to 2, the prbol is the top function nd the cubic is the bottom function: 2 x 2 +x x 3 x x 3 y 4 + xy 2 dy dx Now write the M-file for the integrnd. Think of the expressions in x s representing prmeters: function z=myfun2(x) n=length(x); z=zeros(size(x)); for j=1:n =x(j)^3-x(j); b=x(j)^2+x(j); c=x(j)^3; d=x(j); h=@(y)(c*y.^4+d*y.^2); z(j)=qud(h,,b); In the commnd window, type Q=qud(@myfun2,,2); Exmple: Triple Integrl Here is n exmple problem: Evlute y x 2 x 2 x2 + z y x 2 dz dy dx 2 4
5 (This is sidewys prboloid (y xis running through the top ), cut off t y = 4. For future reference, this could be evluted exctly using polr coordintes: 2π 2 (4 r 2 )r r dr dθ = 128π 15 Using the built-in triple integrl To cll Mtlb s triplequd, your function must be ble to hndle vector x input, nd sclrs y, z. In this exmple, we hve the following: function A=tripletrilfunc(x,y,z) %Assume vector x nd sclrs y nd z % %See if (x,y,z) is in the correct region. If not, return. N=length(x); for j=1:n flg=; if x(j)<=2 & x(j)>=-2 if y>=x(j)^2 & y<=4 if bs(z)<=sqrt(y-x(j)^2) flg=1; if flg== A(j)=; else A(j)=sqrt(x(j).^2+z.^2); Output: >> tic; A=triplequd(@tripletrilfunc,-2,2,-4,4,-4,4); toc Elpsed time is seconds. 5
6 This took LONG time to run- Probbly due to the triple loop nd the requirement to check to see if (x, y, z) is in the region of integrtion. As n lterntive, we cn simply cll qud three times. Notice tht, with fixed vlue of x =, the inside two integrls become: 4 y z y 2 dz dy 2 Furthermore, if y ws fixed prmeter, b, then the inside integrl would simply be: b 2 b z 2 dz This is implemented by the pir of functions below, sved s middle.m: function B=middle(x) %Evlutes the middle integrl, then clls the inside function N=length(x); for j=1:n y1=x(j).^2; y2=4; =x(j); h=@(y) Inside(y,); B(j)=qud(h,y1,y2); function A=Inside(y,c) % Evlutes the inside integrl c=c.^2; %Coming from the outside prmeter (this is fixed x) N=length(y); for j=1:n b=sqrt(y(j)-c); =-b; h=@(z)sqrt(c+z.^2); A(j)=qud(h,,b); In the Mtlb commnd window, we integrte the middle integrl with respect to x (nd the rest of the integrtion is done by our two functions). HINT: First type 6
7 wrning off becuse of the singulrities in our function. >> tic; toc Elpsed time is seconds. By comprison, here re the numericl vlues (Mtlb s defult tolernce is set to 1 6 ): >> Exct=128*pi/ >> Exct-A1 %Using triplequd e-5 >> Exct-A2 %Using the custom three quds e-5 7
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