What fraction of the Earth's volume lies above the 45 degrees north parallel? You may assume the Earth is a perfect sphere. The volume in question is the smaller piece that we would get if the sphere were sliced into two pieces by a plane.
Solution 1: Without loss of generality, look at a sphere of radius 1 centered at the origin. If you like cartesian coordinates, then you can slice the sphere into discs with the same $z$ coordinate, which have radius $\sqrt{1-z^{2}}$, so the region we are considering has volume $\int_{\sqrt{2} / 2}^{1} \pi\left(1-z^{2}\right) d z=\pi\left(\frac{2}{3}-\frac{5 \sqrt{2}}{12}\right)$, and dividing by $4 \pi / 3$ we get $\frac{8-5 \sqrt{2}}{16}$.

Solution 2: For those who prefer spherical coordinates, we can find the volume of the spherical cap plus a cone whose vertex is the center of the sphere. This region is where $r$ ranges from 0 to $1, \theta$ ranges from 0 to $2 \pi$, and $\phi$ ranges from 0 to $\pi / 4$. Remembering we need to subtract off the volume of the cone, which has height $\frac{1}{\sqrt{2}}$ and a circular base of radius $\frac{1}{\sqrt{2}}$, then divide by $\frac{4}{3} \pi$ to get the fraction of the volume of the sphere, we find that we need to evaluate $\frac{\int_{0}^{1} \int_{0}^{2 \pi} \int_{0}^{\pi / 4} r^{2} \sin \phi d \phi d \theta d r-\frac{1}{3} \pi\left(\frac{1}{\sqrt{2}}\right)^{2} \frac{1}{\sqrt{2}}}{4 \pi / 3}$. The integral is just $2 \pi \frac{1}{3}(-\cos \pi / 4+\cos 0)=\frac{4 \pi-2 \pi \sqrt{2}}{6}$. Putting this back in the answer and simplifying yields $\frac{8-5 \sqrt{2}}{16}$.

Solution 3: Cavalieri's Principle states that if two solids have the same cross-sectional areas at every height, then they have the same volume. This principle is very familiar in the plane, where we know that the area of a triangle depends only on the base and height, not the precise position of the apex. To apply it to a sphere, consider a cylinder with radius 1 and height 1 . Now cut out a cone whose base is the upper circle of the cylinder and whose apex is the center of the lower circle. Then at a height $z$ the area is $\pi\left(1^{2}-z^{2}\right)$, exactly the same as for the upper hemisphere! The portion lying above the the 45 degrees north parallel is that which ranges from height $\frac{1}{\sqrt{2}}$ to 1 . The volume of the cylinder in this range is $\pi \cdot 1^{2}\left(1-\frac{1}{\sqrt{2}}\right)$. The volume of the cone in this range is the volume of the entire cone minus the portion from height 0 to $\frac{1}{\sqrt{2}}$, i.e., $\frac{1}{3} \pi\left(1^{2} \cdot 1-\left(\frac{1}{\sqrt{2}}\right)^{2} \frac{1}{\sqrt{2}}\right)$. Therefore the fraction of the Earth's volume that lies above the 45 degrees north parallel is $\frac{\pi\left(1-\frac{1}{\sqrt{2}}\right)-\frac{1}{3} \pi\left(1-\frac{1}{2 \sqrt{2}}\right)}{4 \pi / 3}=\frac{\mathbf{8 - 5} \sqrt{\mathbf{2}}}{\mathbf{1 6}}$.

Solution 4: Another way to approach this problem is to integrate the function $\sqrt{1-x^{2}-y^{2}}$ over the region $x^{2}+y^{2} \leq \frac{1}{\sqrt{2}}$, subtract off a cylinder of radius and height $\frac{1}{\sqrt{2}}$, then divide by the volume of the sphere. One could also use the nontrivial fact that the surface area of a portion of a sphere of radius $r$ between two parallel planes separated by a distance $z$ is $2 \pi r^{2} z$, so in particular the surface area of this cap is $2 \pi\left(1-\frac{1}{\sqrt{2}}\right)$. Now, the ratio of the surface area of the cap to the surface area of the sphere is the same as the ratio of the volume of the cap plus the cone considered in Solution 2 to the volume of the whole sphere, so this allows us to avoid integration entirely.