diff --git a/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.pdf b/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.pdf
index dadcc04..f1bb357 100644
Binary files a/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.pdf and b/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.pdf differ
diff --git a/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.tex b/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.tex
index 0a47b8f..22ddd11 100644
--- a/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.tex
+++ b/advanced-math/exercise/6-differential-calculus-of-multivariate-functions/differential-calculus-of-multivariate-functions.tex
@@ -244,12 +244,30 @@ $\therefore\lim\limits_{(x,y)\to(0,0)}f(x,y)-f(0,0)=bx+cy+o(\rho)$。
 
 \subsubsection{显函数}
 
-首先对原式分别对$xy$求导令其为0，得到极值点。计算二阶微分判断点是否为极值点和为哪种极值点，最后得到极值。
+首先对原式分别对$xy$求导令其为0，得到极值点。利用根的规则计算二阶微分判断点是否为极值点和为哪种极值点，最后得到极值。
 
 \subsubsection{隐函数}
 
 首先对原式分别对$xy$求导，然后令$z_x'$、$z_y'$全部为0得到关系式，再把关系式带回原式得到可疑点。计算二阶微分判断点是否为极值点和为哪种极值点，最后得到极值。
 
+\textbf{例题：}已知对于$z=z(x,y)>0$由$x^2+y^2+z^2-2x-2y-4z-10=0$确定，求其极值。
+
+解：由于$x,y$具有对称性，所以分别求偏导得到：$\dfrac{\partial z}{\partial x}=\dfrac{1-x}{z-2}$，$\dfrac{\partial z}{\partial y}=\dfrac{1-y}{z-2}$。
+
+令偏导数等于0，则得到唯一驻点$(1,1)$。
+
+带入方程解的$(z-6)(z+2)=0$，解得$z(1,1)=6$（$z>0$）。
+
+为判断极值点，需要求二阶偏导。
+
+$\dfrac{\partial^2z}{\partial x^2}=\dfrac{-(z-2)-(1-x)\dfrac{\partial z}{\partial x}}{(z-2)^2}$，所以带入$(1,1)$和$\dfrac{\partial z}{\partial x}=0$，得到$A=-\dfrac{1}{4}$。
+
+$\dfrac{\partial^2z}{\partial x\partial y}=\dfrac{(x-1)\dfrac{\partial z}{\partial y}}{(z-2)^2}$，同理得$B=0$。
+
+$\dfrac{\partial^2z}{\partial y^2}=\dfrac{-(z-2)-(1-y)\dfrac{\partial z}{\partial y}}{(z-2)^2}$，同理得$C=-\dfrac{1}{4}$。
+
+$\Delta<0$且$A>0$，所以得到极大值6。
+
 \subsection{有条件极值}
 
 与无条件极值一样，在边界就是显函数可以直接求，在区域内就是隐函数需要求出可疑点再计算可疑点的二阶导数值判断。
diff --git a/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf b/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf
index 5063fb1..272c6c5 100644
Binary files a/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf and b/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.pdf differ
diff --git a/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex b/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
index 1f120c0..c22a257 100644
--- a/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
+++ b/advanced-math/exercise/7-integral-calculus-of-multivariate-functions/integral-calculus-of-multivariate-functions.tex
@@ -54,31 +54,34 @@
 
 解：对于极坐标的积分次序交换需要利用直角坐标系来画图了解，特别是对于$r$的上下限。
 
-对$\theta=\dfrac{\pi}{2}$变为$y$轴，$y=-\dfrac{\pi}{4}$变为$y=-x$。
+对$\theta=\dfrac{\pi}{2}$变为$y$轴，$\theta=-\dfrac{\pi}{4}$变为$y=-x$。
 
 对$r=2\cos\theta$变为$xy$的表达式，$r^2=2\cos\theta$，即$x^2+y^2=2x$，$(x-1)^2+y^2=1$。
 
+所以所得到的$\sigma$为一个圆割去一个扇形。
+
 \begin{minipage}{0.625\linewidth}
-    所以所得到的$\sigma$为一个圆割去一个扇形。
     
     交换积分次序后就需要以一个长度以极点为圆心做圆，切割$\sigma$。
 
     由$\sigma$可知取长度$\sqrt{2}$可以切分。
+
+    所以$\sigma$可以分为左边的$\sigma_1$和右边的$\sigma_2$。
 \end{minipage}
 \hfill
 \begin{minipage}{0.25\linewidth}
     \begin{tikzpicture}[scale=1]
+        \draw [black, thick, smooth, domain=0:1,fill=gray!20] (1,-1) arc (-90:180:1);
+        \draw (0,0) -- (1,-1);
+        \draw (0,-1.4) arc (-90:90:1.4);
         \draw[-latex](-0.25,0) -- (2.5,0) node[below]{$x$};
         \draw[-latex](0,-1.5) -- (0,1.5) node[above]{$y$};
         \filldraw[black] (0,0) node[below]{$O$};
-        \draw (1,-1) arc (-90:180:1);
-        \draw (0,0) -- (1,-1);
-        \draw (0,-1.4) arc (-90:90:1.4);
+        \filldraw (0.75,0) node{$\sigma_1$};
+        \filldraw (1.65,0) node{$\sigma_2$};
     \end{tikzpicture}
 \end{minipage}
 
-所以$\sigma$可以分为左边的$\sigma_1$和右边的$\sigma_2$。
-
 $\sigma_1$的$r\in[0,\sqrt{2}]$，$\sigma_2$的$r\in[\sqrt{2},2]$。
 
 $\sigma_1$的$\theta$下限是$y=-x$这条边，即$\theta=-\dfrac{\pi}{4}$，上限是$r=2\cos\theta$这个圆，则$\theta=\arccos\dfrac{r}{2}$。