Having some difficulty with a Trip. Integral in Cylindrical Coordinates

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SUMMARY

The discussion focuses on evaluating a triple integral in cylindrical coordinates, specifically the integral of x dV over the region E defined by the planes z=0 and z=x+y+3, and the cylinders x² + y² = 4 and x² + y² = 9. The participants confirm that the limits for r are from 2 to 3 and for theta from 0 to 2π. The upper limit for z is correctly expressed as rcos(θ) + rsin(θ) + 3. The integration process is clarified, leading to a simplified result of (65/4)∫(cos²(θ)) dθ, demonstrating that the integral, while tedious, is manageable.

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Divergent13
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Hi everybody


The integral in question is the triple integral of x dV over the region E, where E is enclosed by the planes z=0, and z=x+y+3, and the cylinders x^2 + y^2 = 4 and x^2 + y^2 = 9.

Well--- so far in cylindrical coordinates I know the r limits will be from 2 to 3 since the cylinders are in the form x^2 + y^2 = r^2

And the Theta limits will be 0 to 2pi.

The z limits are what are bothering me. I believe the lower z limit will be 0, but the upper one is quite confusing. x+y+3 ... am I correct in assuming this should be written as rcos(theta) + rsin(theta) + 3 ??

Lets just say that's right for now (which i know it isn't :frown: ) then I would end up getting an integrand with stuff like cos^2(x) which I know isn't a difficult integral if you use half angles, but it just doesn't seem like it should be this long and difficult. What can I do to change my limits?

Thanks for you help.
 
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There are plenty of us who would say that even with those limits the integral is easy - it can be done by hand, thus it is easy (tedious, long and uninteresting perhaps).
Your analysis seems correct - not all integrals have neat tricks that make things drop out without having to do some genuine integration. but let's see what happens:

\int_{r=2}^3 \int_{\theta=0}^{2\pi}\int_{z=0}^{r\cos\theta+r\sin\theta+3} r^2\cos\theta dzdrd\theta
= \int_{r=2}^3\int_{\theta=0}^{2\pi}r^3\cos^2\theta+r^3\sin\theta\cos\theta+3r^2cos\theta drd\theta

note that the cos(\theta)sin(\theta) integral and the cos(\theta) vanishes, and we can do the r integration first to get

\frac{65}{4}\int_{\theta=0}^{2\pi} cos^2\theta

which is fairly elementary (remember your Fourier series?)
 
Last edited:
Hey thanks matt I believe I got it--- and no that integral is not at all difficult...

Btw I have not studied Fourier series--- only power and taylor series, and just basic ones in the my Calculus III course. :frown:
 
ah, when you come to do Fourier series, you'll have to learn about integrals of cos and sin , and see that they almost always come to zero.
 

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