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 $$ L = \begin{bmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ \frac{1}{2} & 0 & 1 \end{bmatrix} $$ $$ D = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 2 \end{bmatrix} $$...