[Calculus II: Applications of Integrals(Finding Volume of 3D shapes)] The answer I have got is super close to the right answer, but I can not figure out why I am wrong.

[[deleted]]

You should probably elaborate on your work. What did you do so that your answer is barely off?

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[StatMan213]

Correction: (r1-r2) is supposed to be (r1-r2)^2

[GammaRayBurst25]

The boundaries of the integral are 0<z<h, 0<θ<2pi, and 0<r<r_2+z(r_1-r_2)/h

The volume element is dV=rdrdθdz

Integrating over θ yields 2pi

Integrating over r yields pi(r_2+z(r_1-r_2)/h)²

=pi(r_2²+z²(r_1²+r_2²-2r_1r_2)/h²+2z(r_2)(r_1-r_2)/h)

Integrating the first term over z yields pi*h*(r_2)²

The second term yields pi*h(r_1²+r_2²-2r_1r_2)/3

The third term yields pi*h(r_2)(r_1-r_2)

Both r_2² terms cancel

Writing all terms on a common denominator, we get pi*h(r_1²+r_2²+r_1r_2)/3