# 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 $z=1$?

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.

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.

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 $$r= \sqrt{10-1^2}=3$$.

Lower limit in integration for $$r$$ is 0

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