Question

Consider the following. C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

a. Find a piecewise smooth parametrization of the path C.
r(t) = { 0
b Evaluate

Integral of (x+2y^1/2)ds

Answer 1

a.

`\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}`

`\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}`

`\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}`

b.

`\mathbf{\int \limits _(c) F \ dr =(11 √(2)+11)/(6)}`

Explanation:

Given that:

C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1), starting at (0, 0)

a. Find a piecewise smooth parametrization of the path C.

r(t) = { 0

If C: counterclockwise around the triangle with vertices (0, 0), (1, 0), and (0, 1),

Then:

`C_1 = (0,0) \\ \\ C_2 = (1,0) \\ \\ C_3 = (0,1)`

Also:

`\mathtt{r_1 = (0,0) + t(1,0) = (t,0) }`

`\mathbf{r_1 = (t,0) \implies t = 0 \ to \ 1}`

`\mathtt{r_2 = (1,0) + t(-1,1) = (1- t,t) }`

`\mathbf{r_2 = (2-t,t-1) \implies t = 1 \ to \ 2}`

`\mathtt{r_3 = (0,1) + t(0,-1) = (0,1-t) }`

`\mathbf{r_3 = (0,3-t) \implies t = 2 \ to \ 3}`

b Evaluate :

Integral of (x+2y^1/2)ds

`\mathtt{\int \limits ^1_(c1) (x+ 2 √(y)) ds = \int \limits ^1_(0) \ (t + 0) √(1) } \\ \\ \mathtt{ \int \limits ^1_(c1) (x+ 2 √(y)) ds = \begin {pmatrix} (t^2)/(2) \end {pmatrix} }^1_0 \\ \\ \mathtt{\int \limits ^1_(c1) (x+ 2 √(y)) ds = (1)/(2)}`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = \int \limits (x+2 √(y) \sqrt{((dx)/(dt))^2 + ((dy)/(dt))^2 \ dt } }`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = \int \limits 2- t + 2√(t-1) \ √(1+1) }`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = √(2) \int \limits^2_1 2- t + 2√(t-1) \ dt }`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = √(2) } \ \begin {pmatrix} 2t - (t^2)/(2)+ (2(t-1)^(3/2))/(3) (2) \end {pmatrix} ^2_1}`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = √(2) } \ \begin {pmatrix} 2 -(1)/(2) (4-1)+(4)/(3) (1)^(3/2) -0 \end {pmatrix} }`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = √(2) } \ \begin {pmatrix} 2 -(3)/(2) + (4)/(3) \end {pmatrix} }`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = √(2) } \ \begin {pmatrix} (12-9+8)/(6) \end {pmatrix} }`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = √(2) } \ \begin {pmatrix} (11)/(6) \end {pmatrix} }`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = ( √(2) )/(6) \ (11 )}`

`\mathtt{\int \limits _(c2) (x+ 2 √(y)) ds = ( 11 √(2) )/(6)}`

`\mathtt{\int \limits _(c3) (x+ 2 √(y)) ds = \int \limits ^3_2 0+2 √(3-t) \ √(0+1) }`

`\mathtt{\int \limits _(c3) (x+ 2 √(y)) ds = \int \limits ^3_2 2 √(3-t) \ dt}`

`\mathtt{\int \limits _(c3) (x+ 2 √(y)) ds = \int \limits^3_2 \begin {pmatrix} (-2(3-t)^(3/2))/(3) (2) \end {pmatrix}^3_2 }`

`\mathtt{\int \limits _(c3) (x+ 2 √(y)) ds = -(4)/(3) [(0)-(1)]}`

`\mathtt{\int \limits _(c3) (x+ 2 √(y)) ds = -(4)/(3) [-(1)]}`

`\mathtt{\int \limits _(c3) (x+ 2 √(y)) ds = (4)/(3)}`

`\mathtt{\int \limits _(c) F \ dr =(11 √(2))/(6)+(1)/(2)+ (4)/(3)}`

`\mathtt{\int \limits _(c) F \ dr =(11 √(2)+3+8)/(6)}`

`\mathbf{\int \limits _(c) F \ dr =(11 √(2)+11)/(6)}`