Svar Övning 3.3.1
SamverkanFlervariabelanalysLIU
(Skillnad mellan versioner)
Rad 1: | Rad 1: | ||
- | a) <math>D_f=\{(x,y)\in\mathbb{R}^2:\, (x-1)^2+(y-2)^2< | + | a) <math>D_f=\{(x,y)\in\mathbb{R}^2:\, (x-1)^2+(y-2)^2\le2\}</math>. <math>V_f=[0,\sqrt{2}]</math>. |
- | b) <math>D_f=\{(x,y)\in\mathbb{R}^2:\, -1\le x+2y\le 1\}</math>. | + | b) <math>D_f=\{(x,y)\in\mathbb{R}^2:\, -1\le x+2y\le 1\}</math>. <math>V_f=[-\frac{\pi}{2},\frac{\pi}{2}]</math>. |
- | c) <math>D_f=\{(x,y,z)\in\mathbb{R}^3:\, xyz> 0\}</math>. | + | c) <math>D_f=\{(x,y,z)\in\mathbb{R}^3:\, xyz> 0\}</math>. <math>V_f=]-\infty,\infty[</math>. |
Nuvarande version
a) \displaystyle D_f=\{(x,y)\in\mathbb{R}^2:\, (x-1)^2+(y-2)^2\le2\}. \displaystyle V_f=[0,\sqrt{2}].
b) \displaystyle D_f=\{(x,y)\in\mathbb{R}^2:\, -1\le x+2y\le 1\}. \displaystyle V_f=[-\frac{\pi}{2},\frac{\pi}{2}].
c) \displaystyle D_f=\{(x,y,z)\in\mathbb{R}^3:\, xyz> 0\}. \displaystyle V_f=]-\infty,\infty[.