Question

如何用matlab求函数积分的上限b，下限为0

定积分l=∫(x^2+y^2+z^2)^(1/2)*xd(t)上限为b下限为0;其中x=-R2*sin(t);y=R2*cos(t);x=R2*cos(t)/a+(a^2+1)^(1/2)*cos(t)*sin(t)/(a*(R1^2-R2^2*cos(t)^2);已知l=10；R2=5;R1=10;积分下限为0；求上限值。求哪位大神给个答案吧
z=-R2*sin(t);y=R2*cos(t);x=R2*cos(t)/a+(a^2+1)^(1/2)*cos(t)*sin(t)/(a*(R1^2-R2^2*cos(t)^2);

Answer 1

x到底是多少？

追问

z=-R2*sin(t);y=R2*cos(t);x=R2*cos(t)/a+(a^2+1)^(1/2)*cos(t)*sin(t)/(a*(R1^2-R2^2*cos(t)^2);