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 $z=3\sqrt{x^2+y^2}$.
Triple integral. Spherical coordinates
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0$z=\sqrt{3(x^2+y^2)}$ or $z=3\sqrt{(x^2+y^2)}$? – 2012-12-01
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0z= 3 * sqrt(x^2+y^2) – 2012-12-01
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0your second one, sorry I don't know how to type sqrt. – 2012-12-01
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0Please, try to write in TeX. Use dollar signs `$` to enclose your formulas and backslash for commands. For example, `\sqrt{}`. – 2012-12-01
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0For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/) or [here](http://meta.stackexchange.com/a/70559/155238). – 2012-12-01
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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0Hi 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 – 2012-12-02
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0Yep, my mistake, editing now... – 2012-12-02
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0hi your answer is correct. thank you – 2012-12-02
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0but 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 – 2012-12-02
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0The cone extends out infinitely, not stopping once it hits the sphere, so it eventually "surrounds" the top part too. [This](http://ars.els-cdn.com/content/image/1-s2.0-S016521250000055X-gr1.gif) may help visualize. – 2012-12-02
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0ok . thank you so much ! – 2012-12-02
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0@AAbc If you find this answer helpful, please consider upvote it, and accept it: http://meta.math.stackexchange.com/questions/3286/how-do-i-accept-an-answer – 2013-08-02