contestada

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector parametric equation r⃗ (t)r→(t) for the line segment cc so that points pp and qq correspond to t=0t=0 and t=1t=1, respectively. r⃗ (t)=r→(t)= (b) using the parametrization in part (a), the line integral of f⃗ f→ along cc is ∫cf⃗ ⋅dr⃗ =∫baf⃗ (r⃗ (t))⋅r⃗ ′(t)dt=∫ba∫cf→⋅dr→=∫abf→(r→(t))⋅r→′(t)dt=∫ab dtdt with limits of integration a=a= and b=b= (c) evaluate the line integral in part (b). (d) what is the line integral of f⃗ f→ around the clockwise-oriented triangle with corners at the origin, pp, and qq? hint: sketch the vector field and the triangle.

Respuesta :

LammettHash

a. Parameterize [tex]C[/tex] by

[tex]\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath[/tex]

with [tex]0\le t\le1[/tex].

b/c. The line integral of [tex]\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath[/tex] over [tex]C[/tex] is

[tex]\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt[/tex]

[tex]=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt[/tex]

[tex]=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt[/tex]

[tex]=\displaystyle10\int_0^1\mathrm dt=\boxed{10}[/tex]

d. Notice that we can write the line integral as

[tex]\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)[/tex]

By Green's theorem, the line integral is equivalent to

[tex]\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy[/tex]

where [tex]D[/tex] is the triangle bounded by [tex]C[/tex], and this integral is simply twice the area of [tex]D[/tex]. [tex]D[/tex] is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.