Find the value of the line integral. f · dr c (hint: if f is conservative, the integration may be easier on an alternative path.) f(x, y, z) = y sin(z)i + x sin(z)j + xy cos(x)k (a) r1(t) = t2i + t2j, 0 ≤ t ≤ 2 (b) r2(t) = 4ti + 4tj, 0 ≤ t ≤ 1

a) Fdr = 0 b) Fdr = 0

Step-by-step explanation:

According to the integral Fdr

a) If r=t²i + t²j

If the field F(x, y, z) = y sin(z)i + x sin(z)j + xy cos(x)k

Integral Fdr = y sin(z)i + x sin(z)j + xy cos(x)k d(t²i + t²j)

Multiplying component wise.

Fdr = y sin(z)i(dt²i) + x sin(z)j(dt²j) + xy cos(x)k d(0k)

Remember that dot product of different components gives 'zero' i.e i.j = 0 etc.

Fdr = ysin(z)dt² + xsin(z)dt² + 0...(1)

Generally for 2dimensional object r=xi+yj

Comparing this to r = t²i+t²j

x=t² and y = t² z = 0...(2)

Substituting 2 into 1

Fdr = t²sin0dt² + t²sin0dt²

Integrating resulting Fdr with respect to t² within the limit 0 ≤ t ≤ 2, we have

Fdr = 0

b) since there is no z-component in the second part as well, integral of Fdr within the limit 0 ≤ t ≤ 1 will also be 0.