Applying Green's Theorem

The problem asks for the value of the line integral $\oint_C P\,dx + Q\,dy$ around a positively oriented circle $C$. We can use Green's Theorem, which states that:

$ \oint_C P\,dx + Q\,dy = \iint_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA $

where $D$ is the region enclosed by the curve $C$.

Identify P and Q

From the given integral $\oint_C(3y-e^{\cos x})dx+(7x + \sqrt{y^4 +11})dy$, we have:

• $P(x, y) = 3y - e^{\cos x}$
• $Q(x, y) = 7x + \sqrt{y^4 + 11}$

Calculate Partial Derivatives

Next, we compute the partial derivatives:

• $ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3y - e^{\cos x}) = 3 $
• $ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(7x + \sqrt{y^4 + 11}) = 7 $

Compute Integrand Difference

The difference $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ is:

$ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 7 - 3 = 4 $

Evaluate the Double Integral

The curve $C$ is the circle $x^2 + y^2 = 9$, which has a radius $r=3$. The region $D$ enclosed by $C$ is a disk with radius 3. The integral becomes:

$ \oint_C(3y-e^{\cos x})dx+(7x + \sqrt{y^4 +11})dy = \iint_D 4 \, dA $

$ = 4 \iint_D dA $

The term $\iint_D dA$ represents the area of the disk $D$. The area of a circle with radius $r$ is $A = \pi r^2$. For $r=3$, the area is $A = \pi (3^2) = 9\pi$.

Therefore, the value of the double integral is $4 \times (9\pi) = 36\pi$.

Final Calculation

The question asks for the value of $\frac{1}{\pi}\oint_C(3y-e^{\cos x})dx+(7x + \sqrt{y^4 +11})dy$.

$ \frac{1}{\pi} (36\pi) = 36 $

The value of the expression is 36.