beginner integration trouble.

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)

    2.7493

Area=2*int(f,0,sol);
disp(['Area under the curve f(x) is: ',char(Area)]);

Area under the curve f(x) is: 2*int((x^(3/2) - x^2 + 3)^(1/2), x, 0, 3095424455315773/1125899906842624)

vol=int(pi*(f)^2,x,a,sol);

Unrecognized function or variable 'a'.

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.

[Walter Roberson]

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.

Answer 1

[Torsten]

"int" cannot find an analytical antiderivative for f(x).

Thus to get the numerical value for the integral, use

Area=2*vpaintegral(f,0,sol);

instead of

Area=2*int(f,0,sol);