Understanding the Integral Regions
The integral $I$ is given as the sum of two integrals over distinct regions, $R_1$ and $R_2$. Our goal is to combine these regions into a single region $R = R_1 \cup R_2$ and express the integral $I$ by changing the order of integration from $dx \, dy$ to $dy \, dx$.
Region 1 ($R_1$) Analysis
The first integral is $I_1 = \int_{0}^{1} \int_{0}^{\sqrt{y}} dx \, dy$. The region $R_1$ is defined by the bounds:
- Outer bounds: $0 \le y \le 1$
- Inner bounds: $0 \le x \le \sqrt{y}$
The condition $x \le \sqrt{y}$ implies $x^2 \le y$ because $x$ is non-negative. Considering the range of $y$ ($0 \le y \le 1$), the maximum value for $\sqrt{y}$ is $1$. Thus, the bounds for $x$ are $0 \le x \le 1$. Rewriting $R_1$ for integration order $dy \, dx$:
$R_1$: $0 \le x \le 1$, $x^2 \le y \le 1$
Region 2 ($R_2$) Analysis
The second integral is $I_2 = \int_{1}^{2} \int_{\sqrt{y-1}}^{1} dx \, dy$. The region $R_2$ is defined by the bounds:
- Outer bounds: $1 \le y \le 2$
- Inner bounds: $\sqrt{y-1} \le x \le 1$
The condition $x \ge \sqrt{y-1}$ implies $x^2 \ge y-1$ (since $x \ge 0$), which rearranges to $y \le x^2+1$. As $y$ ranges from $1$ to $2$, $\sqrt{y-1}$ ranges from $0$ to $1$. Given the inner bound $x \le 1$, the overall range for $x$ is $0 \le x \le 1$. Rewriting $R_2$ for integration order $dy \, dx$:
$R_2$: $0 \le x \le 1$, $1 \le y \le x^2+1$
Combining Regions for Changed Order
To change the order of integration for the total integral $I = I_1 + I_2$, we consider the combined region $R = R_1 \cup R_2$. The range of $x$ over $R$ is the union of the $x$-ranges for $R_1$ and $R_2$, which is $[0, 1]$.
Now, we determine the bounds for $y$ for a fixed $x \in [0, 1]$:
- In $R_1$, $y$ ranges from $x^2$ up to $1$.
- In $R_2$, $y$ ranges from $1$ up to $x^2+1$.
Combining these, for any fixed $x$ in $[0, 1]$, $y$ varies continuously from the minimum lower bound ($x^2$) to the maximum upper bound ($x^2+1$).
Therefore, the bounds for the integral with the order changed ($dy \, dx$) are:
- Outer bounds: $0 \le x \le 1$
- Inner bounds: $x^2 \le y \le x^2+1$
Final Integral Expression
The integral $I$ with the order of integration changed is represented as:
$I = \int_{0}^{1} \int_{x^2}^{x^2+1} dy \, dx$Important Questions from Area Under Curve
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 toLet $C: x^2 + y^2 =9$ 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_____.
