# Thread: Double Integrals in Polor Coordinates

1. ## Double Integrals in Polor Coordinates

Above z = sqrt(x^2 + y^2) and below the sphere x^2 + y^2 + z^2 =1

Any help is appreciated.

2. Originally Posted by JonathanEyoon 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.

3. 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.

4. Originally Posted by JonathanEyoon 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.
Correct.

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.
They meet when $\displaystyle \sqrt{1-(x^2+y^2)}=\sqrt{x^2+y^2}\implies 1-(x^2+y^2)=x^2+y^2\implies 1=2x^2+2y^2$ $\displaystyle \implies x^2+y^2=\tfrac{1}{2}$

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?

5. Originally Posted by Chris L T521 Correct.

They meet when $\displaystyle \sqrt{1-(x^2+y^2)}=\sqrt{x^2+y^2}\implies 1-(x^2+y^2)=x^2+y^2\implies 1=2x^2+2y^2$ $\displaystyle \implies x^2+y^2=\tfrac{1}{2}$

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?

Ok I worked it out myself and I arrived at the same answer you did for the radius. Appreciated! Now my next question is why is it the sphere minus the cone in the integrand? Why can it not be the other way around?

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