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I need to calculate the following in cylindrical coordinates:
$$\iiint_K \sqrt{x^2+y^2+z^2}\,dx\,dy\,dz$$ $K$ is bounded by the plane $z=3$ and by the cone $x^2+y^2=z^2$.

I know that:
$x=r\cos{\theta}$
$y=r\sin{\theta}$
$z=z$

So I get the following: $$\iiint r\sqrt{r^2+z^2}\,d\theta\, dr\,dz$$

The problem I have is finding from where to where I need to integrate.
I tried the following which was wrong:
$$\int_0^{2\pi}\,d\theta \int_0^3\,dz \int_0^z r\sqrt{r^2+z^2}\,dr$$

This gave me the following result: $\frac{24}{5}\sqrt{3}\pi$
Which is wrong according to my coursebook.

Shaun
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Joshua
  • 529

2 Answers2

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I would say that the limits of integration are correct. Let's recalculate:

$\int_0^{2\pi}\,d\theta \int_0^3\,dz \int_0^z r\sqrt{r^2+z^2}\,dr=2\pi \int_0^3\,dz \int_0^z r\sqrt{r^2+z^2}\,dr=\frac{2}{3}\pi\int_0^3[\sqrt{(r^2+z^2)^3}]_0^z\,dz=$

$=\frac{2(2\sqrt2-1)}{3}\pi\int_0^3z^3\,dz=\frac{(2\sqrt2-1)}{6}\pi[z^4]_0^3=\frac{27(2\sqrt2-1)}{2}\pi$

What the coursebook say such an outcome?

georg
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1

You have the right process, the problem is not worded correctly (The equation for the solid is incorrect). If $x^2+y^2=z^2$ then it's true that $$z=\sqrt{x^2+y^2}\\z=\sqrt{r^2} = r$$ If you adjust for this correction your process will be correct.

cdosborn
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