diff --git a/semester3/analysis-ii/cheat-sheet-jh/analysis-ii-cheat-sheet.pdf b/semester3/analysis-ii/cheat-sheet-jh/analysis-ii-cheat-sheet.pdf
index f760283..2a4034d 100644
Binary files a/semester3/analysis-ii/cheat-sheet-jh/analysis-ii-cheat-sheet.pdf and b/semester3/analysis-ii/cheat-sheet-jh/analysis-ii-cheat-sheet.pdf differ
diff --git a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/01_int_in_rn.tex b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/01_int_in_rn.tex
index acf35c8..a69ec6e 100644
--- a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/01_int_in_rn.tex
+++ b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/01_int_in_rn.tex
@@ -67,6 +67,9 @@ Image of par. curve $\gamma: [a, b] \rightarrow \R^n$ is negligible, since $\gam
 
 \rmvspace
 \bi{How to compute the integral:} We compute each integral "inside out". For a definite integral, don't just find the anti-derivative, compute the actual integral!
-For an integral as seen in the harder example, we compute it as we normally would, simply using the $\pm 2x$ as the $a$ and $b$
+For an integral as seen in the harder example, we compute it as we normally would, simply using the $\pm 2x$ as the $a$ and $b$.
+
+Using a change of variables into polar coordinates may come in handy, e.g. for a set like $\{ (x, y) \in \R^2 \divides 1 \leq x^2 + y^2 \leq 4 \}$,
+we can use polar coordinates and the integral is then $\int_{0}^{2\pi} \int_{1}^{2} f(x, y) \dx r \dx \varphi$ (or flipped of course)
 
 \rmvspace
diff --git a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex
index 7e65f8c..01d839f 100644
--- a/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex
+++ b/semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex
@@ -53,4 +53,4 @@ Be cognizant of what direction the integral goes, if the set is on the right han
 If the curve doesn't fully enclose the set, then we can simply compute the line integrals of the missing sections and subtract them from the final result.
 
 \shade{gray}{Center of mass}
-% TODO: Finish the notes here
+The center of mass of an object $\cU$ is given by $\displaystyle \overline{x}_i = \frac{1}{\text{Vol}(\cU)} \int_{\cU} x_i \dx x$.