Question

verify that \\( mathbf { f } ( x, y ) = left( 6 x y ^ { 3 } + 3
ight) mathbf { i } + left( 9 x ^ { 2 } y ^ { 2 } + 2 e ^ { 2 y }
ight) mathbf { j } \\) is conservative and find a potential function \\( f \\) for \\( mathbf { f } \\). then use \\( f \\) and the fundamental theorem of line integrals to evaluate \\( int _ { c } mathbf { f } cdot d mathbf { r } \\), where \\( c \\) is the curve given by \\( mathbf { r } ( t ) = left( 2 sin ^ { 5 } t
ight) mathbf { i } + left( \frac { 2 t - sin ( 5 t ) } { 12 }
ight) mathbf { j } , 0 leq t leq pi / 2 \\). \\( f ( x , y ) = \\) \\( int _ { c } mathbf { f } cdot d mathbf { r } = \\)

Explanation:

Step1: Verify F is conservative

Check if $\frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y}$.
$F_x = 6xy^3 + 3$, so $\frac{\partial F_x}{\partial y} = 18xy^2$.
$F_y = 9x^2y^2 + 2e^{2y}$, so $\frac{\partial F_y}{\partial x} = 18xy^2$. Equal, so conservative.

Step2: Find potential function f

Integrate $F_x$ with respect to x:
$f(x,y) = \int (6xy^3 + 3)dx = 3x^2y^3 + 3x + g(y)$.

Step3: Determine g(y)

Differentiate f w.r.t y: $\frac{\partial f}{\partial y} = 9x^2y^2 + g'(y) = F_y = 9x^2y^2 + 2e^{2y}$.
Thus $g'(y)=2e^{2y}$, integrate: $g(y)=e^{2y}+C$. Take C=0: $f=3x^2y^3 + 3x + e^{2y}$.

Step4: Find curve endpoints

At $t=0$: $\mathbf{r}(0)=(0)\mathbf{i} + (0)\mathbf{j}=(0,0)$.
At $t=\pi/2$: $\mathbf{r}(\pi/2)=(2\sin^5(\pi/2))\mathbf{i} + (\frac{2(\pi/2)}{\pi}\sin(12\cdot5\cdot\pi/2))\mathbf{j}=(2, \sin(30\pi))=(2,0)$.

Step5: Apply Fundamental Theorem

$\int_C \mathbf{F}\cdot d\mathbf{r} = f(2,0) - f(0,0)$.
$f(2,0)=3(4)(0)+3(2)+e^0=6+1=7$.
$f(0,0)=0+0+1=1$.
Difference: $7-1=6$.

Answer:

6