# Triple integral. Spherical coordinates

Find the volume of the solid that lies within the sphere $x^2 + y^2 + z^2 =25$, above the $xy$-plane, and outside the cone $z=3\sqrt{x^2+y^2}$.

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$z=\sqrt{3(x^2+y^2)}$ or $z=3\sqrt{(x^2+y^2)}$? –  Sigur Dec 1 '12 at 23:44
z= 3 * sqrt(x^2+y^2) –  AAbc Dec 1 '12 at 23:49
your second one, sorry I don't know how to type sqrt. –  AAbc Dec 1 '12 at 23:50
Please, try to write in TeX. Use dollar signs $ to enclose your formulas and backslash for commands. For example, \sqrt{}. – Sigur Dec 1 '12 at 23:50 For some basic information about writing math at this site see e.g. here or here. – Américo Tavares Dec 1 '12 at 23:54 ## 2 Answers Hints: $$z=3\sqrt{x^2+y^2}\Longrightarrow 25=x^2+y^2+9x^2+9y^2\Longrightarrow x^2+y^2=2.5\Longrightarrow$$ the cone interesects the sphere on the above rightmost circle. Since everything symetric with respect all the axis and the origin, you can try to calcualte the volumet in the first octant and the multiply by 4 (as we're interested only in what happens above the$\,xy-$plane. Finally, you can try to calculate the volume between the$\,xy-$plane and inside the cone (that is inside the sphere): $$4\int_0^{\sqrt{2.5}}\int_0^{\sqrt{2.5-x^2}}\int_0^{3\sqrt{x^2+y^2}}dzdydx\,\,\,\\\text{Very strongly adviced to change this to cylindrical coordinates}$$ and then substract this from the half sphere's volume. $${}$$ Disclaimer: The above are only general hints. Check it thoroughly. - Using the following substitutions for spherical coordinates: $$z = \rho \cos(\phi)$$ $$x = \rho \sin(\phi)\cos(\theta)$$ $$y = \rho \sin(\phi)\sin(\theta)$$ We can substitute these into your equations: $$\rho ^2 < 25 \implies \rho < 5 \\ \text{Above the xy -plane} \implies \phi > \pi / 2 \\ z < 3 \sqrt{x^2 + y^2} \implies z^2 < 9(x^2 + y^2) \implies (\rho \cos(\phi))^2 < 9(\rho \sin(\phi))^2 \implies \tan(\phi) < 1/3$$$\phi < \tan^{-1}(1/3) \approx 20.48^o\$

Now we can set up our triple integral:

$$\int_0^{2\pi} \int_{20.48}^{90} \int_0^5 \rho^2 \sin(\phi) d\rho d\phi d\theta$$

Inner: $$1/3 \rho^3 \sin(\phi) ]_0^5 = 125/3 sin(\phi)$$ Outer: $$-125/3\cos(\phi)]_{20.48}^{90} = -125/3(0-0.9487) = 39.529$$ Outer(last): $$39.529]_0^{2 \pi} = 248.369$$

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Hi this is what I was trying to do. but would't this be the cone only? since phi is from 0 to 20.48. but i want area under the cone . thank you –  AAbc Dec 2 '12 at 0:28
Yep, my mistake, editing now... –  Ben Dec 2 '12 at 0:29
hi your answer is correct. thank you –  AAbc Dec 2 '12 at 0:43
but i don't understand why is it 20.48 to 90. this would be the volume of the semi sphere cut out the cone form 20.48 to 90. but on top the cone, wouldn't there be a little part as well. since the cone is flat at the top and sphere is round. by going form 20.48-90 we ignore the little part from 0 to 20.48 –  AAbc Dec 2 '12 at 0:46
The cone extends out infinitely, not stopping once it hits the sphere, so it eventually "surrounds" the top part too. This may help visualize. –  Ben Dec 2 '12 at 1:17