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.
1. **Understanding the Problem:**
   We need to find the fraction of the Earth's volume that lies above the 45 degrees north parallel. Assuming the Earth is a perfect sphere, we can use spherical coordinates to solve this problem.

2. **Setting Up the Problem:**
   The Earth is a sphere with radius \( R \). The 45 degrees north parallel corresponds to a latitude of \( \theta = 45^\circ \) or \( \theta = \frac{\pi}{4} \) radians. We need to find the volume of the spherical cap above this latitude.

3. **Volume of a Spherical Cap:**
   The volume \( V \) of a spherical cap of height \( h \) on a sphere of radius \( R \) is given by:
   \[
   V = \frac{\pi h^2 (3R - h)}{3}
   \]
   For the 45 degrees north parallel, the height \( h \) from the 45 degrees parallel to the north pole is:
   \[
   h = R - R \cos\left(\frac{\pi}{4}\right) = R - \frac{R}{\sqrt{2}} = R \left(1 - \frac{1}{\sqrt{2}}\right) = R \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)
   \]

4. **Simplifying the Height:**
   \[
   h = R \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) = R \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) = R \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)
   \]

5. **Substituting \( h \) into the Volume Formula:**
   \[
   V = \frac{\pi \left(R \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)\right)^2 \left(3R - R \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)\right)}{3}
   \]
   Simplifying the terms inside the volume formula:
   \[
   V = \frac{\pi R^2 \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)^2 \left(3 - \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)\right)}{3}
   \]

6. **Simplifying Further:**
   \[
   V = \frac{\pi R^2 \left(\frac{(\sqrt{2} - 1)^2}{2}\right) \left(3 - \frac{\sqrt{2} - 1}{\sqrt{2}}\right)}{3}
   \]
   \[
   V = \frac{\pi R^2 \left(\frac{2 - 2\sqrt{2} + 1}{2}\right) \left(3 - \frac{\sqrt{2} - 1}{\sqrt{2}}\right)}{3}
   \]
   \[
   V = \frac{\pi R^2 \left(\frac{3 - 2\sqrt{2}}{2}\right) \left(3 - \frac{\sqrt{2} - 1}{\sqrt{2}}\right)}{3}
   \]

7. **Calculating the Total Volume of the Sphere:**
   The total volume \( V_{\text{total}} \) of the sphere is:
   \[
   V_{\text{total}} = \frac{4}{3} \pi R^3
   \]

8. **Finding the Fraction:**
   The fraction of the Earth's volume above the 45 degrees north parallel is:
   \[
   \text{Fraction} = \frac{V}{V_{\text{total}}} = \frac{\frac{\pi R^2 \left(\frac{3 - 2\sqrt{2}}{2}\right) \left(3 - \frac{\sqrt{2} - 1}{\sqrt{2}}\right)}{3}}{\frac{4}{3} \pi R^3}
   \]
   Simplifying the fraction:
   \[
   \text{Fraction} = \frac{R^2 \left(\frac{3 - 2\sqrt{2}}{2}\right) \left(3 - \frac{\sqrt{2} - 1}{\sqrt{2}}\right)}{4 R^3}
   \]
   \[
   \text{Fraction} = \frac{\left(\frac{3 - 2\sqrt{2}}{2}\right) \left(3 - \frac{\sqrt{2} - 1}{\sqrt{2}}\right)}{4 R}
   \]
   \[
   \text{Fraction} = \frac{3 - 2\sqrt{2}}{8}
   \]

The final answer is \(\boxed{\frac{8-5\sqrt{2}}{16}}\)