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Homework20.tex
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Homework20.tex
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\documentclass{scrartcl}
\usepackage{amsmath,amssymb,commath}
\setkomafont{disposition}{\normalfont\bfseries}
\newcommand{\Ln}{\text{Ln}}
\newcommand{\Arg}{\text{Arg}}
\newcommand{\Rp}{\text{Re}}
\title{Complex Analysis}
\subtitle{Homework 20: 5.1) 14, 28, 32}
\author{Kenny Roffo}
\date{Due November 18, 2015}
\begin{document}
\maketitle
\textbf{14)} Evaluate the line integrals $\int_CG(x,y)dx, \int_CG(x,y)dy$ and $\int_CG(x,y)ds$ on the curve
$$G(x,y)\frac{x^2}{y^3}; \text{\hspace{0.5in}} 2y=3x^{3/2} \text{\hspace{0.5in}} 1\le x\le8$$
\begin{align*}
\int_CG(x,y)dx &= \int_1^8\left(\frac{x^2}{y^3}\right)dx\\
&= \int_1^8\left(\frac{x^2}{\left(\frac{3}{2}x^{3/2}\right)^3}\right)dx\\
&= \frac{3^3}{2^3}\int_1^8\left(\frac{x^2}{x^{9/2}}\right)dx\\
&= \frac{8}{27}\int_1^8x^{2-9/2}dx\\
&= \frac{8}{27}\int_1^8x^{-5/2}dx\\
&= \frac{8}{27}\left[-\frac{2}{3}x^{-3/2}\right]_1^8\\
&= \frac{16}{81}\left[-8^{-3/2} + 1^{-3/2}\right]\\
&= \frac{16}{81}\left[1 - \frac{1}{8\sqrt{8}}\right]\\
&\approx 0.1888
\end{align*}\pagebreak
\begin{align*}
\int_CG(x,y)dy &= \int_1^8\frac{x^2}{y^3}\od{}{x}\left(\frac{3}{2}x^{3/2}\right)dx\\
&= \int_1^8\frac{x^2}{\left(\frac{3}{2}x^{3/2}\right)^3}x^{1/2}dx\\
&= \frac{8}{27}\int_1^8x^{2+1/2-9/2}dx\\
&= \frac{8}{27}\int_1^8x^{-2}dx\\
&= \frac{8}{27}\left[-x^-1\right]_1^8\\
&= \frac{8}{27}\left[1 - \frac{1}{8}\right]\\
&\approx 0.259
\end{align*}
\begin{align*}
\int_CG(x,y)ds &= \int_1^8\frac{x^2}{\left(\frac{3}{2}x^{3/2}\right)^3}\sqrt{1+\left[\od{}{x}\left(\frac{3}{2}x^{3/2}\right)\right]^2}dx\\
&= \int_1^8\frac{x^2}{\left(\frac{3}{2}x^{3/2}\right)^3}\sqrt{1+\left[x^{1/2}\right]^2}dx\\
&= \frac{8}{27}\int_1^8\frac{x^2}{x^{9/2}}\sqrt{1+x}dx\\
&= \frac{8}{27}\left[-\frac{2}{3}\frac{(x+1)^{3/2}}{x^{3/2}}\right]_1^8 \text{\hspace{1in} via WolframAlpha.com}\\
&= 0.323
\end{align*}\pagebreak
\textbf{28)} Evaluate $\oint_C(x^2+y^2)dx - 2xydy$ on the closed curve $C = C_1 \cup C_2$ where $$C_1 = \left\{(x,y) | y=x^2, 0 \le x \le 1\right\} \text{\hspace{0.5in} and \hspace{0.5in}} C_2 = \left\{(x,y) | y=\sqrt{x}, 0 \le x \le 1\right\}$$
\begin{align*}
\oint_C \left(x^2+y^2\right)dx - 2xydy &= \oint_{C_1} \left(x^2+y^2\right)dx - 2xydy + \oint_{C_2} \left(x^2+y^2\right)dx - 2xydy\\
&= \oint_0^1 \left(x^2+\left(x^2\right)^2\right)dx - 2x(x^2)(2xdx) + \oint_0^1 (x^2+(\sqrt{x})^2)dx - 2x\sqrt{x}\left(\frac{1}{2}x^{-1/2}dx\right)\\
&= \oint_0^1 \left(x^2+x^4\right) - 4x^4 + \left(x^2+x\right) - x\ dx\\
&= \oint_0^1 -3x^4 + 2x^2\ dx \\
&= \left[-\frac{3}{5}x^5 + \frac{2}{3}x^2\right]_0^1\\
&= \frac{2}{3} - \frac{3}{5}\\
&= \frac{19}{15}\\
&\approx 3.267
\end{align*}\pagebreak
\textbf{32)} Evaluate $\int_{-C}ydx - xdy$ where $C$ is given by $x = 2\cos t, y=3\sin t, 0\le t\le \pi$
\begin{align*}
\int_{-C}ydx - xdy &= \int_\pi^0\left(3\sin t\right)(-2\sin t)dt - \left(2\cos t\right)(3\cos t)dt\\
&= \int_\pi^0-6\sin^2t - 6\cos^2tdt\\
&= -6\int_\pi^01dt\\
&= 6\pi
\end{align*}
\end{document}