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

Answer: 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.