Above z = sqrt(x^2 + y^2) and below the sphere x^2 + y^2 + z^2 =1
Any help is appreciated.
To find the region R, note that $\displaystyle \sqrt{1-(x^2+y^2)}=\sqrt{x^2+y^2}\implies x^2+y^2=\tfrac{1}{2}$
Thus,
$\displaystyle 0\leq r\leq \tfrac{\sqrt{2}}{2}$
$\displaystyle 0\leq\vartheta\leq2\pi$
$\displaystyle r\leq z\leq \sqrt{1-r^2}$
Thus, your volume integral is $\displaystyle \int_0^{2\pi}\int_0^{\frac{\sqrt{2}}{2}}\left(\sqr t{1-r^2}-r\right)r\,dr\,d\vartheta=\int_0^{2\pi}\int_0^{\fr ac{\sqrt{2}}{2}}r\sqrt{1-r^2}\,dr\,d\vartheta-\int_0^{2\pi}\int_0^{\frac{\sqrt{2}}{2}}r^2\,dr\,d \vartheta$
Does this make sense?
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As a side note, if you were dealing with a triple integral (in cylindrical coordinates), you would be evaluating $\displaystyle \int_0^{2\pi}\int_0^{\frac{\sqrt{2}}{2}}\int_r^{\s qrt{1-r^2}}r\,dz\,dr\,d\vartheta$ which yields the two double integrals I gave you above.
I'm unclear about how you were able to get the radius limits. Can you explain a little more what you did in that first line to get you the radius? I was told the radius limits would be where the cone meets the sphere. Usually to find points of intersection we set two functions equal to each other and solve but in this case I don't know what I'm doing. Thanks.
Correct.
They meet when $\displaystyle \sqrt{1-(x^2+y^2)}=\sqrt{x^2+y^2}\implies 1-(x^2+y^2)=x^2+y^2\implies 1=2x^2+2y^2$ $\displaystyle \implies x^2+y^2=\tfrac{1}{2}$Usually to find points of intersection we set two functions equal to each other and solve but in this case I don't know what I'm doing. Thanks.
The region that is created by the intersection of the two surfaces is a circle with radius $\displaystyle \tfrac{1}{\sqrt{2}}$
Thus, the radius is defined on the interval $\displaystyle 0\leq r\leq \tfrac{1}{\sqrt{2}}$
Does this make sense?