The question is : For the solid bounded by the xy plane the cylinder $x^2+y^2=a^2$, the paraboloid $z=b(x^2+y^2)$ with $b>0$, find the centroid. I have found the volume which is checked to be true. $$V = \int_0^{2\pi}\int_0^abr^2rdrd\theta= \frac{1}{2}\pi a^4b$$ Then I let the density be 1 so that mass is equal to volume. $$V_z = \int_0^{2\pi}\int_0^a z^2br^2rdrd\theta=\int_0^{2\pi}\int_0^ab^2r^5drd\theta = \frac{1}{3}\pi a^6b^2$$ $$\overline{z} = \frac{V_z}{V}=\frac{\frac{1}{3}\pi a^6b^2}{\frac{1}{2}\pi a^4b}=\frac{2}{3}a^2b$$ While the true answer is $\frac{1}{3}a^2b$ I know there is a method using triple integral to find the centroid, but this question is designed for double integral exercise. It comes from Calculus with Analytic Geometry second edition P.730 Q31 Can you help me or state where do i got a mistake?

  • $\begingroup$ $V_z=\int_0^{2\pi}\int_0^{a}\frac{1}{2}z^2rdrd\theta$ because $\int zdz=\frac{1}{2}z^2+c$. $\endgroup$
    – Bob Dobbs
    Dec 13, 2022 at 11:45


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