$\sqrt{{\left(\frac{d}{\mathrm{dt}}\left(3\+2 \cos \left(t\right)\right)\right)}^2\+{\left(\frac{d}{\mathrm{dt}}\left(5\+2 \sin \left(t\right)\right)\right)}^2}$

$\=\sqrt{4\left({\sin }^2\left(t\right)\+{\cos }^2\left(t\right)\right)}$

$\=2$

Initialize

Loading Student:-VectorCalculus

Access the PathInt command through the Context Panel

$x y$$\stackrel{\text{line integral}}{\to }$${\∫}_0^{2\mathrm{\π}}2\left(3\+2\cos \left(t\right)\right)\left(5\+2\sin \left(t\right)\right)\sqrt{{\sin \left(t\right)}^2\+{\cos \left(t\right)}^2}\ⅆt$$\=$$60\mathrm{\π}$

Form and evaluate the line integral via the PathInt command

$\mathrm{PathInt}\left(x y\,\left[x\,y\right]\=\mathrm{Circle}\left(⟨3\,5⟩\,2\right)\,\mathrm{output}\=\mathrm{integral}\right)$

${\∫}_0^{2\mathrm{\π}}2\left(3\+2\cos \left(t\right)\right)\left(5\+2\sin \left(t\right)\right)\sqrt{{\sin \left(t\right)}^2\+{\cos \left(t\right)}^2}\ⅆt$

$\mathrm{PathInt}\left(x y\,\left[x\,y\right]\=\mathrm{Circle}\left(⟨3\,5⟩\,2\right)\right)$ = $60\mathrm{\π}$$$

Define $f$ as a scalar-valued function of a vector argument

$f\left(\mathbf{v}\right)\=v_1\cdot v_2$$\stackrel{\text{assign as function}}{\to }$$f$

Define the circle parametrically as the position vector R

$\mathbf{R}\=⟨3\+2 \cos \left(t\right)\,5\+2 \sin \left(t\right)⟩$$\stackrel{\text{assign}}{\to }$

Obtain $\mathrm{\ρ}\=\frac{\mathrm{ds}}{\mathrm{dt}}\=∥\stackrel{\.}{\mathbf{R}}∥$ 

$∥\frac{\ⅆ}{\ⅆ t} \mathbf{R}∥$ = $2$$$

Form and evaluate the line integral ${\int }_0^{2 \mathrm{\π}}f\left(\mathbf{R}\right) \mathrm{\ρ} \mathrm{dt}$ 

$2{\int }_0^{2 \mathrm{\π}}f\left(\mathbf{R}\right) \mathit{\ⅆ}t$ = $60\mathrm{\π}$ $$

Explicit formulation and evaluation of the line integral

$2 f\left(\mathbf{R}\right)$

$2\left(3\+2\cos \left(t\right)\right)\left(5\+2\sin \left(t\right)\right)$

$\stackrel{\text{integrate w.r.t. t}}{\to }$

${\∫}_0^{2\mathrm{\π}}2\left(3\+2\cos \left(t\right)\right)\left(5\+2\sin \left(t\right)\right)\ⅆt$

$\=$

$60\mathrm{\π}$