Let $C: x^2 + y^2 =9$ be the circle in $R^2$ oriented positively. Then $\frac{1}{\pi}\oint_C(3y-e^{\cos x})dx+(7x + \sqrt{y^4 +11})dy$ equals_____.
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.
Important Questions from Area Under Curve
- Let $I$ be the integral defined as follows: $$I = \int_{0}^{1} \int_{0}^{\sqrt{y}} dx dy + \int_{1}^{2} \int_{\sqrt{y-1}}^{1} dx dy$$ If the order of the integration is changed, then which one of the following is the correct expression for $I$?
Let $\alpha = \iint_S \vec{F} \cdot \hat{n} \, dS$, where $\vec{F} = (2x + 3z)\hat{i} + (xz - y)\hat{j} + (y^2 + 2z)\hat{k}$ and $S$ is the sphere with centre at $(3, -1, 2)$ and radius 9. Here, $\hat{n}$ is the unit normal drawn outward and $\hat{i}, \hat{j}, \hat{k}$ are unit vectors.
Then the value of $\frac{1}{36\pi} \alpha$ is equal to ________. (answer in integer)
The value of $\frac{4}{\pi} \int_0^{\pi/2} \sin^2 x \text{ dx}$ is _________________ (rounded off to two decimal places).
The work done by the force $F = (x + y)\hat{i} - (x^2 + y^2)\hat{j}$, where $\hat{i}$ and $\hat{j}$ are unit vectors in $\vec{OX}$ and $\vec{OY}$ directions, respectively, along the upper half of the circle $x^2 + y^2 = 1$ from $(1,0)$ to $(-1,0)$ in the $xy$-plane is
If the line $y = \alpha x$, $\alpha \geq \sqrt{2}$, divides the area of the region
$R: = \{(x, y) \in \mathbb{R}^2| 0 \leq x \leq \sqrt{y}, 0 \leq y \leq 2\}$
into two equal parts, then the value of $\alpha$ is equal to
