Question Number 134599 by mohammad17 last updated on 05/Mar/21
Answered by Olaf last updated on 05/Mar/21
![a) ∫_C ∣z∣^2 dz = ∫_0 ^(√2) ∣re^(i(π/4)) ∣^2 dr = [(r^3 /3)]_0 ^(√2) = ((2(√2))/3) b) ∫_C ∣z∣dz = ∫_0 ^(√2) ∣re^(i(π/4)) ∣dr = [(r^2 /2)]_0 ^(√2) = 1 c) ∫_C xdz = ∫_0 ^(√2) rcos(π/4)dr = [(r^2 /(2(√2)))]_0 ^(√2) = (1/( (√2)))](https://www.tinkutara.com/question/Q134600.png)
$$\left.{a}\right) \\ $$$$\int_{\mathrm{C}} \mid{z}\mid^{\mathrm{2}} {dz}\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \mid{re}^{{i}\frac{\pi}{\mathrm{4}}} \mid^{\mathrm{2}} {dr}\:=\:\left[\frac{{r}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:=\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$$\left.{b}\right) \\ $$$$\int_{\mathrm{C}} \mid{z}\mid{dz}\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \mid{re}^{{i}\frac{\pi}{\mathrm{4}}} \mid{dr}\:=\:\left[\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:=\:\mathrm{1} \\ $$$$\left.{c}\right) \\ $$$$\int_{\mathrm{C}} {xdz}\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} {r}\mathrm{cos}\frac{\pi}{\mathrm{4}}{dr}\:=\:\left[\frac{{r}^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$