beginner integration trouble.
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
Aryaman
am 4 Okt. 2024
Kommentiert: Walter Roberson
am 5 Okt. 2024
CODE:
```
syms x
f(x)= (x^(3/2)+3-x^2)^(1/2);
g(x)= -(x^(3/2)+3-x^2)^(1/2);
sol=double(solve(f==0));
sol=sol(sol==real(sol));
disp(sol)
Area=2*int(f,0,sol);
disp(['Area under the curve f(x) is: ',char(Area)]);
vol=int(pi*(f)^2,x,a,sol);
disp(['volume of solid of rotation formed by the curve f(x) and g(x) about x axis is: ',char(vol)]);
%```
sol=sol(sol==real(sol)) %, this part will remove any imaginary terms stored in sol.
output:
>> DA2_Q1_b
2.749289201023484
Area under the curve f(x) is: 2*int((x^(3/2) - x^2 + 3)^(1/2), x, 0, 3095424455315773/1125899906842624)
volume of solid of rotation formed by the curve f(x) and g(x) about x axis is: (3095424455315773*pi*(623191256382180861935616*3095424455315773^(1/2) + 9136014217435573277565931304275))/21408715390589398215874289541742427045741199360
as you can see the output for area is not what i wanted. Can anyone please help me to get a numeric or symbolic answer to area.
1 Kommentar
Walter Roberson
am 5 Okt. 2024
Maple is able to produce an exact solution for
2*int((x^(3/2) - x^2 + 3)^(1/2), x, 0, 3095424455315773/1125899906842624)
However, it is a long expression that involves a lot of occurances of expressions similar to
(RootOf(_Z^4 - _Z^3 - 3, index = 1) - RootOf(_Z^4 - _Z^3 - 3, index = 2))*RootOf(_Z^4 - _Z^3 - 3, index = 4)/((RootOf(_Z^4 - _Z^3 - 3, index = 1) - RootOf(_Z^4 - _Z^3 - 3, index = 4))*RootOf(_Z^4 - _Z^3 - 3, index = 2))
Those can be converted to closed form radicals, but then the expressions involve a lot of things such as
(-2*(12 + 4*sqrt(265))^(2/3) + (12 + 4*sqrt(265))^(1/3) + 32)/(12 + 4*sqrt(265))^(1/3)
You can get an exact solution using Maple, but it is a pretty messy exact solution -- the kind of solution that reading it leaves you less enlightened than not reading it.
Akzeptierte Antwort
Weitere Antworten (0)
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!