Ideas on calculating the volume of $K:=\{(x,y,z)\in \mathbb R^{3}: x^2+y^2+z^2\leq 1, x^2+y^2\leq a\}$ Calculate the volume of $K:=\{(x,y,z)\in \mathbb R^{3}: x^2+y^2+z^2\leq 1, x^2+y^2\leq a\}$ and note that $0 < a < 1$
My ideas: 
depending on the size $a$, I have a cylinder whose height is restricted by the radius of the unit sphere.
$\lambda^{3}(K)=\int_{K}d\lambda^{3}=\int_{[0,2\pi]}\int_{[0,a]}\int_{[-\sqrt{1-a},\sqrt{1-a}]}rdzdrd\phi=4\pi \sqrt{1-a}\int_{[0,a]}rdr=4\pi\sqrt{1-a}\frac{1}{2}a^2=2\pi a^2 \sqrt{1-a}$
Is this correct?
Note: if $a = 1$, I will use $\lambda^{2}$ and in this case $\lambda^{2}(K)=\frac{4}{3}\sqrt{\pi}$

This is a solid of revolution so can be found using the cylindrical shell method
$$2\int_0^{\sqrt{a}}2\pi rh\,dx$$
where $r=x$ and $h=\sqrt{1-x^2}$
Note that this is an elementary integral and you will get the result
$$ V=\frac{4\pi}{3}\left[1-(1-a)^{3/2}\right] $$
