63.3k views
3 votes
If C is the line segment from 7 3 to 0 0 find the value of the line integral C 4y2i xj dr

User Alsein
by
7.9k points

1 Answer

3 votes
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=-\frac{189}2
User Mnestorov
by
7.8k points