Integrate f(x, y, z) = (x 2 + y 2 )0.5 over the involute curve, r(t) = (cost+tsint) i + (sint– tcost) j, 0 ≤ t ≤ 3.

The problem involves a line integral over a vector field. However, the function f(x, y, z) = (x^2 + y^2)^0.5 is a scalar field, not a vector field. The curve r(t) = (cos(t) + tsin(t))i + (sin(t) - tcos(t))j is a parameterization of a curve in the xy-plane.

The line integral of a scalar field f over a curve parameterized by r(t), a ≤ t ≤ b, is given by the formula:

∫_a^b f(r(t)) ||r'(t)|| dt

Here, ||r'(t)|| is the magnitude of the derivative of r with respect to t.

First, we need to find r'(t). Differentiating the components of r with respect to t gives:

r'(t) = (-sin(t) + sin(t) + tcos(t))i + (cos(t) + cos(t) + tsin(t))j = (tcos(t))i + (2cos(t) + t*sin(t))j

The magnitude of r'(t) is then:

||r'(t)|| = sqrt[(tcos(t))^2 + (2cos(t) + t*sin(t))^2]

Next, we substitute r(t) into f to get f(r(t)):

f(r(t)) = ((cos(t) + tsin(t))^2 + (sin(t) - tcos(t))^2)^0.5

Finally, we substitute these into the line integral formula and evaluate the integral from t = 0 to t = 3. This will give the value of the line integral of f over the curve r.

Please note that the actual computation of the integral might be quite complicated and may require numerical methods.