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I am trying to calculate the mass of the object surrounded by the surface $\,\,\,x^2+\frac{y^2}{4}+\frac{z^2}{9}=1$ where the density is give by $\mu(x,y,z)=e^\sqrt{x^2+\frac{y^2}{4}+\frac{z^2}{9}}$

if you try to do this using triple integrals you will get a very hard integral, so instead I tried to do it using surface integral as it lowers the number of variables by one, the formula I used is $$M=\iint_B \mu(x,y,f(x,y))\cdot\sqrt{1+(f_x)^2+(f_y)^2}\,dy\,dx$$ where $B$ is the projection of the surface on x0y-plane and $z=f(x,y)=3\sqrt{1-x^2-\frac{y^2}{4}}$ ,but this integral ended being tough too, I tried the Parametrization $x=r\cos\theta,y=2r\sin\theta$ but I failed to complete.

I ran out of tools and I don't know how to complete from here, thank you very much!

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    $\begingroup$ A simple change of variables (scaling) followed by a switch to spherical coordinates will help. $\endgroup$ Jun 10, 2023 at 16:24

2 Answers 2

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Let $(x,y,z)=(u,2v,3w)$ to integrate the mass as $$ M=\iiint_B \mu(x,y,z)dxdydz = 6\iiint_{u^2+v^2+w^2<1}e^{\sqrt{u^2+v^2+w^2}}dudvdw\\ =6\int_0^{2\pi}\int_0^\pi \int_0^1 e^{r}r^2\sin\theta dr d\theta d\phi=24\pi(e-2) $$

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    $\begingroup$ thank you very much $\endgroup$
    – user1187347
    Jun 10, 2023 at 16:50
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If you do the change of variables $X=x$, $Y=\frac y2$, and $Z=\frac z3$, the your integral becomes$$6\iiint_Be^{\sqrt{X^2+Y^2+Z^2}}\,\mathrm dX\,\mathrm dY\,\mathrm dZ,$$where $B$ is the closed unit ball. And now you can use spherical coordinates; the previous integral is equal to\begin{align}6\int_0^{2\pi}\int_0^\pi\int_0^1\rho^2e^\rho\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta&=2\pi\left(\int_0^\pi\sin\varphi\,\mathrm d\varphi\right)\left(\int_0^1\rho^2e^\rho\,\mathrm d\rho\right)\\&=24\pi(e-2).\end{align}

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  • $\begingroup$ the trick I was missing is the change of variables thank you sir $\endgroup$
    – user1187347
    Jun 10, 2023 at 16:50

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