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The question asks me to formulate a triple integral for the volume of the cap of the solid sphere $x^2 + y^2 + z^2 \leq 10$ cut off by the plane $z=1$?

Here's the integral I formulated as an answer (I used cylindrical coordinates):

$$\int_0^{2\pi} \int_0^\sqrt{10} \int_1^\sqrt{10-r^2} r \ dz \ dr \ d\theta$$

However, apparently the correct answer is:

$$\int_0^{2\pi} \int_0^3 \int_1^\sqrt{10-r^2} r \ dz \ dr \ d\theta$$

The only difference is the $3$, but where does the $3$ come from? $x^2 + y^2 + z^2 \leq 10$ implies we're dealing with a sphere with radius $\sqrt{10}$. I have no idea where the $3$ is coming from really. Any help is appreciated.

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When $z=1$ you have

$$x^2+y^2+z^2=10$$ $$x^2+y^2+1^2=10$$ $$x^2+y^2=9$$

... a circle of radius 3. This "shadow" of the solid defines the region in the xy plane the integration is happening over.

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The upper boundary of the radial $r$-integral is determined by

$$r^2= x^2+y^2= 10-z^2$$

Given that the volume is capped at $z=1$, the maximum radius is $r= \sqrt{10-1^2}=3$.

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Lower limit in integration for $r$ is 0

Upper limit in integration for $r$ is $ \sqrt{(\sqrt{10})^2 -1^2 }$

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