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 f399f2b..346931d 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 b42eb75..9b5eac7 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
@@ -141,4 +141,23 @@ $\int_0^\pi\cos^2x\,\textrm{d}x\int_0^{\sin x}\textrm{d}y=\int_0^\pi\cos^2x\sin
 
 所以$=2\int_0^{\frac{\pi}{4}}\textrm{d}\theta\int_0^{\frac{1}{\cos\theta}}\textrm{d}r=2\int_0^{\frac{\pi}{4}}\dfrac{\textrm{d}\theta}{\cos\theta}=2\ln(\sec\theta+\tan\theta)|_0^{\frac{\pi}{4}}=2\ln(1+\sqrt{2})$。
 
+\subsection{二重积分求导}
+
+即对二重积分求导，需要将二重积分化为一重积分。
+
+
+\subsection{二重积分等式}
+
+\textbf{例题：}设$f(x,y)$为连续函数，且$f(x,y)=\dfrac{1}{\pi}\sqrt{x^2+y^2}\iint\limits_{x^2+y^2\leqslant1}f(x,y)\,\textrm{d}\sigma+y^2$，求$f(x,y)$。
+
+解：$\because f(x,y)$为连续函数，所以其在区间上可积且是一个常数。
+
+令$\iint\limits_{x^2+y^2\leqslant1}f(x,y)\,\textrm{d}\sigma=A$。对$f(x,y)=\dfrac{A}{\pi}\sqrt{x^2+y^2}+y^2$两边积分：
+
+$A=\dfrac{A}{\pi}\iint\limits_{x^2+y^2\leqslant1}\sqrt{x^2+y^2}\,\textrm{d}\sigma+\iint\limits_{x^2+y^2\leqslant1}y^2\,\textrm{d}\sigma$，令$x=r\cos\theta$，$y=r\sin\theta$：
+
+$A=\dfrac{A}{\pi}\int_0^{2\pi}\textrm{d}\theta\int_0^1r^2\,\textrm{d}r+\int_0^{2\pi}\sin^2\theta\textrm{d}\theta\int_0^1r^3\,\textrm{d}r=\dfrac{2A}{3}+\dfrac{\pi}{4}$。$A=\dfrac{3}{4}\pi$。
+
+则代入原式$f(x,y)=\dfrac{3}{4}\sqrt{x^2+y^2}+y^2$。
+
 \end{document}