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# Multivariable Calculus, Spherical Coordinates?

Evaluate the triple integral of x^2+y^2 dV where the domain is the solid hemisphere x^2+y^2+z^2 is less than or equal to 9 and y is greater than or equal to 0

*z is greater than or equal to 0.

no restrictions for 7 are given

sorry!

### 1 Answer

- NickLv 68 years agoFavorite Answer
Spherical Polar Coordinates are related to Rectangular Cartesian Coordinates by:

x = rsinθcosφ (1i)

y = rsinθsinφ (1ii)

z = rcosθ (1iii)

so:

x^2 + y^2 = r^2sin^2(θ) (2)

and the infinitesimal volume in Spherical Polar Coordinates is:

dV = r^2sinθ dr dθ dφ (3)

So your integral is seperable:

∫∫∫ r^2sin^2(θ) dV = ∫[0,2π] dφ ∫[0,π/2] sin^3(θ) dθ ∫[0,3] r^4 dr

The limits are for the hemisphere for which θ and φ are in the positive z-region and r is for a sphere radius 3.

Using the trig identity sin^2θ + cos^2θ = 1:

= 2π[(1/5)r^5]_[0,3] ∫[0,π/2] (sin(θ) - sin(θ)cos^2(θ)) dθ

= π(2/5)(3)^5[-cos(θ) + (1/3)cos^3(θ)]_[0,π]

= π(2/5)(3)^5(-(cos(π/2) - cos(0)) + (1/3)(cos^3(π/2) - cos^3(0)))

= π(2/5)(3)^5(-(0 - 1) + (1/3)(0 - 1))

= π(2/5)(3)^5( 1 - (1/3) )

= π(2/5)(3)^5(2/3)

= (2/5)(3)^4(2)

= π(2^2)(3^4)/5 = 203.58

Double check the working, just in case.

Edit: Sorry is right! I just went through that lot again, I think it's right. I have no idea what you mean by: "no restrictions for 7 are given" did you mistype 'y'.