Question

If C is the line segment from 7 3 to 0 0 find the value of the line integral C 4y2i xj dr

Answer 1

Parameterize $C$ by

$\mathbf r(t)=(x(t),y(t))=(1-t)(7,3)+t(0,0)=(7-7t,3-3t)$

where $0\le t\le1$. Then

$\mathrm d\mathbf r=(-7\,\mathbf i-3\,\mathbf j)\,\mathrm dt$

and the line integral is equivalent to

$\displaystyle\int_C(4y^2\,\mathbf i+x\,\mathbf j)\cdot\mathrm d\mathbf r=\int_0^1(4(3-3t)^2\,\mathbf i+(7-7t)\,\mathbf j)\cdot(-7\,\mathbf i-3\,\mathbf j)\,\mathrm dt$
$\displaystyle=\int_0^1(-252t^2+525t-273)\,\mathrm dt=-\dfrac{189}2$