Applying the steps, the result of the line integral is 99.
The curve is:
[tex]f(x,y,z) = xy i + 9y^2 j[/tex]
The vector field is:
[tex]r(t) = (x(t), y(t)) = (16t^6, t^4)[/tex]
Applying the vector field at the curve, we have that:
[tex]f(t) = (16t^{10}, 9t^8)[/tex]
The derivative of the vector field is:
[tex]r^{\prime}(t) = (96t^5, 4t^3)[/tex]
The dot product of the vector field along the curve with the derivative is:
[tex]f(t)r^{\prime}(t) = (16t^{10}, 9t^8)(96t^5, 4t^3) = 1536t^{15} + 36t^{11}[/tex]
Hence, the line integral is:
[tex]I = \int_{0}^{1} (1536t^{15} + 36t^{11}) dt[/tex]
[tex]I = 96t^{16} + 3t^{12}|_{t = 0}^{t = 1}[/tex]
[tex]I = 96 + 3[/tex]
[tex]I = 99[/tex]
A similar problem is given at https://brainly.com/question/12666512
