(z-a)√(R^2-2AZ+A^2)对z的积分,积分域从-R到R

Question

(z-a) √ (R^2-2AZ+A^2)对z的积分,积分域从-R到R

Answer 1

∫ (z-a)√(r^2-2az+a^2)dz
=∫ -1/(2a)[(r^2-2za+a^2)-(r^2-a^2)]√(r^2-2az+a^2)dz
=∫ -1/(2a)[(r^2-2za+a^2)]√(r^2-2az+a^2)dz
-∫ -1/(2a)[(r^2-a^2)]√(r^2-2az+a^2)dz
=∫ 1/(2a)^2*(r^2-2za+a^2)^(3/2)d(r^2-2az+a^2)
-∫ 1/(2a)^2*[(r^2-a^2)]*(r^2-2az+a^2)^(1/2)d(r^2-2az+a^2)
=[ 1/(2a)^2*2/5*(r^2-2za+a^2)^(5/2)]
-[ 1/(2a)^2*(r^2-a^2)*2/3*(r^2-2az+a^2)^(3/2)]
=[1/(2a)^2*2/5*((r-a)^5-(r+a)^5)]
-[1/(2a)^2*(r^2-a^2)*2/3*((r-a)^3-(r+a)^3)]
=[1/(2a)^2*2/5*(-2a)*(5r^4+10r^2a^2+a^4)]
-[1/(2a)^2*(r^2-a^2)*2/3*(-2a)*(3r^2+a^2)]
=[-1/(5a)*(5r^4+10r^2a^2+a^4)]-[-1/(3a)*(r^2-a^2)*(3r^2+a^2)]
=[-1/(5a)*(5r^4+10r^2a^2+a^4)]-[-1/(3a)*(3r^4-2r^2a^2-a^4)]
=1/a*[(-r^4-2r^2a^2-1/5*a^4)+(r^4-2/3*r^2a^2-1/3*a^4)]
=-8a/15*(5r^2+a^2)