Dr. Brajesh Kumar Jha

Applications of Double Integrals Double integration is a fundamental tool in multivariable calculus used to compute volumes, areas, mass, centroids, and physical quantities over two-dimensional regions. 1. Volume Under a Surface If $z = f(x,y)$ over region $R$, the volume is: $$ V = \iint_R f(x,y)\, dA $$ Illustration Surface 2. Area of a Plane Region The area of region $R$: $$ A = \iint_R 1\, dA $$ Example Region (Circle) R 3. Mass of a Lamina If density is $\rho(x,y)$: $$ M = \iint_R \rho(x,y)\, dA $$ If density is constant $\rho = k$: $$ M = k \cdot \text{Area}(R) $$ 4. Center of Mass (Centroid) $$ \bar{x} = \frac{1}{M} \iint_R x \rho(x,y)\, dA $$ $$ \bar{y} = \frac{1}{M} \iint_R y \rho(x,y)\, dA $$ Centroid Illustration (x̄, ȳ) 5. Moments of Inertia $$ I_x = \iint_R y^2 \rho(x,y)\, dA $$ $$ I_y = \iint_R x^2 \rho(x,y)\, dA $$ $$ I_0 = \iint_R (x^2 + y^2)\rho(x,y)\...

Applications of Double Integrals Area Enclosed by Plane Curves Double integration is used to compute area, volume, mass and many physical quantities. $$A = \iint_R 1\, dA$$ Example 1 Find the area of a quadrant of $x^2 + y^2 = a^2$. 🔽 Show Step-by-Step Solution Using vertical strips: $$A=\int_0^a \int_0^{\sqrt{a^2-x^2}} dy\,dx$$ $$A=\int_0^a \sqrt{a^2-x^2}\,dx$$ Using standard integral: $$\int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$$ Evaluating limits: $$A=\frac{\pi a^2}{4}$$ Final Answer: $$\boxed{\frac{\pi a^2}{4}}$$ Example 2 Find the area bounded by $y=2-x$, $y^2=2(x+2)$. 🔽 Show Step-by-Step Solution Intersection points: $$(2-x)^2=2(x+2)$$ $$x=0,6$$ Using horizontal strips: $$A=\int_{-4}^{2}\int_{(y^2/2)-2}^{2-y} dx\,dy$$ $$A=\int_{-4}^{2}\left(4-y-\frac{y^2}{2}\right) dy$$ After evaluation: $$A=18$$ Final Answer: $$\boxed{18}$$ Example 3 Find the a...

LDL T Factorization – Detailed Numerical Examples Dr. Brajesh Kumar Jha General Formula For symmetric matrix $A$: $$ A = L D L^T $$ $$ d_k = a_{kk} - \sum_{j=1}^{k-1} l_{kj}^2 d_j $$ $$ l_{ik} = \frac{1}{d_k} \left( a_{ik} - \sum_{j=1}^{k-1} l_{ij} l_{kj} d_j \right) $$ Example 1 $$ A = \begin{bmatrix} 4 & 2 & 2 \\ 2 & 5 & 1 \\ 2 & 1 & 3 \end{bmatrix} $$ Step 1 $$ d_1 = 4 $$ $$ l_{21} = \frac{2}{4} = \frac{1}{2}, \quad l_{31} = \frac{2}{4} = \frac{1}{2} $$ Step 2 $$ d_2 = 5 - \left(\frac{1}{2}\right)^2(4) $$ $$ = 5 - 1 = 4 $$ $$ l_{32} = \frac{1}{4} \left( 1 - \frac{1}{2}\cdot\frac{1}{2}\cdot 4 \right) = \frac{1}{4}(1-1)=0 $$ Step 3 $$ d_3 = 3 - \left(\frac{1}{2}\right)^2(4) $$ $$ = 3 - 1 = 2 $$ Final Matrices $$ L = \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{2} & 0 & 1 \end{bmatrix} $$ $$ D = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 2 \end{bmatrix} $$...

20 Important Questions on Triple Integration 20 Important Questions on Triple Integration Basic Cartesian Integrals 1. Evaluate: $$\iiint_V (x+y+z)\, dV$$ where $0 \le x \le 1$, $0 \le y \le 2$, $0 \le z \le 3$. 2. Evaluate: $$\iiint_V xyz \, dV$$ where $0 \le x \le 2$, $0 \le y \le 1$, $0 \le z \le 3$. 3. Evaluate: $$\int_0^1 \int_0^{2x} \int_0^{x+y} dz\, dy\, dx$$ 4. Change the order of integration: $$\int_0^1 \int_0^{1-x} \int_0^{1-x-y} f(x,y,z)\, dz\, dy\, dx$$ Region Between Surfaces 5. Evaluate: $$\iiint_V z \, dV$$ where $z = 0$ and $z = 4 - x^2 - y^2$. 6. Find the volume bounded by $$z = x^2 + y^2 \quad \text{and} \quad z = 4$$ 7. Evaluate: $$\iiint_V (x^2+y^2)\, dV$$ inside cylinder $x^2 + y^2 = 9$, $0 \le z \le 5$. 8. Find the mass of the cube $0 \le x,y,z \le 1$ with density $\rho = x+y+z$. Spherical Coordinates 9. Find the volume of sphere $$x^2+y^2+z^2 \le a^2$$ 10. Evaluate: ...

Triple Integration 1. Introduction Let $f(x,y,z)$ be a continuous function defined on a closed and bounded region $V \subset \mathbb{R}^3$. The triple integral of $f$ over $V$ is defined as $$ \iiint_V f(x,y,z)\, dV $$ It represents the limit of Riemann sums: $$ \iiint_V f(x,y,z)\, dV = \lim_{\max \Delta V_i \to 0} \sum f(x_i,y_i,z_i)\,\Delta V_i $$ provided the limit exists. If $f(x,y,z)=1$, then the triple integral reduces to: $$ \iiint_V 1\, dV = \text{Volume of } V $$ In practical computation, triple integrals are evaluated as iterated integrals: $$ \iiint_V f(x,y,z)\, dV = \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{h_1(x,y)}^{h_2(x,y)} f(x,y,z)\, dz\, dy\, dx $$ The order of integration may be changed whenever convenient. # 2. Explanation Think of a double integral as adding up tiny rectangles to find the area of a region. A triple integral does the same thing in three dimensions — it adds up tiny boxes (small volumes) to find: Volume...

Double Integration Examples with Graphs Double Integration: Solved Examples with Graphs Understanding Double Integration Double integration extends the concept of a single integral to functions of two variables, $$f(x, y)$$. Geometrically, it represents the volume under a surface within a given region $$D$$. Example 1: Rectangular Bounds Evaluate $$\iint_R (x + 2y) \, dA$$ over $$R = [0, 1] \times [0, 2]$$. x y Step 1: Setup Iterated Integral $$\int_{0}^{1} \int_{0}^{2} (x + 2y) \, dy \, dx$$ Step 2: Integrate w.r.t $y$ $$\int_{0}^{1} [xy + y^2]_{0}^{2} \, dx = \int_{0}^{1} (2x + 4) \, dx$$ Step 3: Integrate w.r.t $x$ $$[x^2 + 4x]_{0}^{1} = 1 + 4 = 5$$ Final Answer: 5 Example 2: Polar Transformation Find the area of a circle with radius $$a$$. $$\iint_D dA = \int_{0}^{2\pi} \int_{...

What is Weighted Sampling? Weighted sampling is a sampling technique where each item is assigned a weight , and the probability of selecting that item is proportional to its weight. Higher weight $\Rightarrow$ higher chance of being selected Lower weight $\Rightarrow$ lower chance (but usually not zero) Mathematical Definition Given $n$ items with weights: $ w_1, w_2, \dots, w_n $ The probability of selecting item $i$ is: $ P(i) = \frac{w_i}{\sum_{j=1}^{n} w_j} $ Example: weights $[1,3,6] \Rightarrow [0.1,0.3,0.6]$ Visual Intuition Weighted bar representation: A | █ B | ███ C | ██████ Flattened view: | A | B B B | C C C C C C | Sampling is equivalent to picking a random point on this bar. Larger segments correspond to higher probability. Four Examples of Weighted Sampling Lottery: More tickets $\Rightarrow$ higher winning chance Online advertising: Higher bids shown more often Survey sampling: Oversampling rare groups G...