How would you do this?

1. Oct 25, 2013

Superposed_Cat

How would you go about solving this?

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2. Oct 25, 2013

arildno

Integrate it?

3. Oct 25, 2013

UltrafastPED

It's a triple integral (a volume, for example). You need two more sets of integration limits. Since one of the integration variables doesn't appear in the integrand you can integrate it directly - it would most likely end up as a factor of 2pi. Looking at the remaining parts, you can integrate each separately, taking into account the integration limits.

4. Oct 26, 2013

TysonM8

c∫d α∫β 1∫2 ρ^2 sin∅ dρd∅dθ

= c∫d α∫β [(ρ^3)/3 sin∅] [1,2] d∅dθ
= c∫d α∫β 7/3 sin∅ d∅dθ
= c∫d [-7/3 cos∅] [α,β] dθ
= c∫d [-7/3 cosβ] - [-7/3 cosα] dθ
= c∫d 7/3 (cosα - cosβ) dθ
= [(7θ)/3 (cosα - cosβ)] [c,d]
= [(7d)/3 (cosα - cosβ)] - [(7c)/3 (cosα - cosβ)]
= (1/3)(7d-7c)(cosα - cosβ)

This is the general solution where α≤∅≤β and c≤∅≤d

5. Oct 26, 2013

Superposed_Cat

Sorry, cropped too much there.

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6. Oct 26, 2013

HallsofIvy

Staff Emeritus
That's easy then! Since the limits of integraton are all constants and the integrand is just a product of functions of the separate variables, that is just
$$\left(\int_1^2 \rho^2 d\rho\right)\left(\int_{\pi/2}^\pi \sin(\phi) d\phi\right)\left(\int_0^2 d\theta\right)$$

Though I must say the limits of integration on that last, $d\phi$, integral look suspicious to me! It's not impossble- but "0 to 2 radians" seems peculiar. "0 to $2\pi$" would see much more reasonable. That would be integrating over a hemi-spherical "cup", below the xy-plane with thickness from 1 to 2.

7. Oct 26, 2013

Superposed_Cat

Why is that weird?

8. Oct 26, 2013

Crake

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9. Oct 26, 2013

Crake

You're integrating using spherical coordinates. $\theta$ is an angle and it seems weird that it goes from $0$ to $2$ radians.