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Hello I'm trying to compute this triple integral

assume(H>0);
integrate(1,z,sqrt(x^2+y^2),H+sqrt(3)*y/3);
integrate(%,y,-sqrt(H^2*(3-sqrt(3))^2/4-x^2),sqrt(H^2*(3-sqrt(3))^2/4-x^2));
integrate(%,x,-H*(3-sqrt(3))/2,H*(3-sqrt(3))/2);

And there is a variable H. So I want the final result to be a function of H. But I get a different answer from sage and Mathematica. I hope the integrals are the same. What am I doing wrong? The sage output produces complex solutions because of the argument of inverse sine being larger than 1 while the Mathematica output is nice real solution. Which one is correct?

This is the sage output

This is the Mathematica output

Hello I'm trying to compute this triple integral

assume(H>0);
integrate(1,z,sqrt(x^2+y^2),H+sqrt(3)*y/3);
integrate(%,y,-sqrt(H^2*(3-sqrt(3))^2/4-x^2),sqrt(H^2*(3-sqrt(3))^2/4-x^2));
integrate(%,x,-H*(3-sqrt(3))/2,H*(3-sqrt(3))/2);

And there is a variable H. So I want the final result to be a function of H. But I get a different answer from sage and Mathematica. I hope the integrals are the same. What am I doing wrong? The sage output produces complex solutions because of the argument of inverse sine being larger than 1 while the Mathematica output is nice real solution. Which one is correct?

This is the sage output

This is the Mathematica output

Hello I'm trying to compute this triple integral

assume(H>0);
integrate(1,z,sqrt(x^2+y^2),H+sqrt(3)*y/3);
integrate(%,y,-sqrt(H^2*(3-sqrt(3))^2/4-x^2),sqrt(H^2*(3-sqrt(3))^2/4-x^2));
integrate(%,x,-H*(3-sqrt(3))/2,H*(3-sqrt(3))/2);

And there is a variable H. So I want the final result to be a function of H. But I get a different answer from sage and Mathematica. I hope the integrals are the same. What am I doing wrong? The sage output produces complex solutions because of the argument of inverse sine being larger than 1 while the Mathematica output is nice real solution. Which one is correct?

This is the sage output

This is the Mathematica output