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[tex] \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2-y^2}dy [/tex]

I can calculate the above integral [part of a double integral] by the conventional way [somewhat long], however my book says that this integral equals to [tex]\frac {\pi}{2}(r^2-x^2)[/tex] because the integral is actually the area of half a circle. I have difficulty to understand how this integral has something to do with half of a circle. [The author of my book thinks this is fact is trivial, however for me it is not =(]

And if the above integral is the area of half a cicle than what about:

[tex] \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2)dy [/tex]

I'll appreciate any help.

[EDIT: I think now I understand why the first integral is the area of half a cirlce, now thinking what I can tell about the second integral...]

I can calculate the above integral [part of a double integral] by the conventional way [somewhat long], however my book says that this integral equals to [tex]\frac {\pi}{2}(r^2-x^2)[/tex] because the integral is actually the area of half a circle. I have difficulty to understand how this integral has something to do with half of a circle. [The author of my book thinks this is fact is trivial, however for me it is not =(]

And if the above integral is the area of half a cicle than what about:

[tex] \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2)dy [/tex]

I'll appreciate any help.

[EDIT: I think now I understand why the first integral is the area of half a cirlce, now thinking what I can tell about the second integral...]

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