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I'm trying to find the integral of $f(x,y,z)=kz^2$, outside of the cylinder $x^2+y^2=a^2$, and inside the sphere $x^2+y^2+z^2=4a^2$. I got the integral:

$\int_0^{2\pi}\int_0^{a^2}\int_0^{\sqrt{4a^2-r^2}}r dzdrd\theta $

thank you!

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  • $\begingroup$ "The volume of $f$" is again nonsense. The integral of $f$ makes sense. $\endgroup$ – Martín-Blas Pérez Pinilla Mar 4 '18 at 23:14
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Since you are using cylindrical coordinates all you to do is place the term $kz^2$ into the $3^{rd}$ integral. You should also check the way you bound $r,θ$

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It's clear that the cylinder $x^2 + y^2 = a^2$ is bounded by the sphere $x^2+y^2+z^2= 4a^2$.

Therefore to the find the area bounded by the function $f(x,y,z) = kz^2$ inside the sphere and outside the cylinder, you would want to calculate the area bounded by the function inside the sphere, and then take away the area bounded by the function inside the cylinder.

Effectively, this would leave you with the bounded area inside the circle, but not inside the cylinder.

Combining these into one integral would be more challenging, as the integral for the sphere is with respect to $θ , \phi\ ,r$ and the integral for the cylinder is with respect to $z,r,θ$. Therefore I would rethink your method and reconsider your integral.

As for $kz^2$, this would be what you integrate. In your integral you have only used $r$, which I assume is your Jacobian for your change of variables. However, when calculating an area bounded by a function, you must multiply your Jacobian by the function, then integrate.

i.e. for your integral, you would integrate $rkz^2$, not just $r$.

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To many formulas for a comment:

The sphere is: $$z^2 = 4a^2 - (x^2 + y^2) = 4a^2 - r^2,$$ $$z = \pm\sqrt{4a^2 - r^2}.$$

The cylinder is: $$a^2 = x^2 + y^2 = r^2,$$ $$r = a.$$

But "bounded by the graph of $f(x,y,z) = kz^2$" does not makes sense in $\Bbb R^3$ because the graph of $f$ is in $\Bbb R^4$. Being $f$ the integrand makes more sense. $z = f(x,y)$ for some $f$ of two variables also makes sense.

Your integration limits are a total disaster. For example, $r$ from $0$ to $a^2$ is the interior of a cylinder of radius... $a^2$! But you want the exterior of the cylinder $r = a$, i.e., $r \ge a$. For the upper bound of $r$, use the sphere.

Check what is actually $f$ and edit your question with the new integration limits.

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