Application of Double Integration 1

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.

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1. Volume Under a Surface

If $z = f(x,y)$ over region $R$, the volume is:

$$ V = \iint_R f(x,y)\, dA $$

Illustration

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2. Area of a Plane Region

The area of region $R$:

$$ A = \iint_R 1\, dA $$

Example Region (Circle)

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3. Mass of a Lamina

If density is $\rho(x,y)$:

$$ M = \iint_R \rho(x,y)\, dA $$

If density is constant $\rho = k$:

$$ M = k \cdot \text{Area}(R) $$

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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

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5. Moments of Inertia

$$ I_x = \iint_R y^2 \rho(x,y)\, dA $$ $$ I_y = \iint_R x^2 \rho(x,y)\, dA $$ $$ I_0 = \iint_R (x^2 + y^2)\rho(x,y)\, dA $$

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6. Probability Applications

If $f(x,y)$ is a joint probability density function:

$$ P((x,y)\in R) = \iint_R f(x,y)\, dA $$ $$ \iint_{\mathbb{R}^2} f(x,y)\, dA = 1 $$

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7. Surface Area

If $z = f(x,y)$:

$$ A = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dA $$

Surface Illustration

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End of Notes

Applications of Double Integrals

Area Enclosed by Plane Curves

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Illustration 16.1

Find the area of a quadrant of the circle

$$x^2 + y^2 = a^2$$

Solution:

$$ A=\int_{0}^{a}\int_{0}^{\sqrt{a^2-x^2}} dy\,dx = \int_{0}^{a} \sqrt{a^2-x^2}\, dx $$

$$ A=\frac{\pi a^2}{4} $$

Answer: $$\boxed{\frac{\pi a^2}{4}}$$

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Illustration 16.2

Find the area bounded by

$$y = 2-x, \qquad y^2 = 2(x+2)$$

Intersection:

$$ (2-x)^2=2(x+2) $$ $$ x^2-6x=0 \Rightarrow x=0,6 $$

Using horizontal strips: $$ A=\int_{-4}^{2}\left(4-y-\frac{y^2}{2}\right)dy $$

$$ A=18 $$

Answer: $$\boxed{18}$$

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Illustration 16.3

Find the area bounded by

$$2x-3y+4=0,\quad x+y-3=0,\quad y=0$$

Intersection Points:

$$(-2,0),\; (3,0),\; (1,2)$$

$$ A=\int_{0}^{2}\int_{\frac{3y-4}{2}}^{3-y} dx\,dy $$

$$ A=5 $$

Answer: $$\boxed{5}$$

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Illustration 16.4

Find the area bounded by

$$x^2=4(y+2), \quad x^2=3-y$$

Intersection:

$$y=-1, \quad x=\pm 2$$

$$ A=\int_{-2}^{2}\left(5-\frac{5x^2}{4}\right)dx $$

$$ A=\frac{172}{15} $$

Answer: $$\boxed{\frac{172}{15}}$$

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Illustration 16.5

Find the area bounded by

$$x(x^2+y^2)=a(x^2-y^2)$$

Using polar substitution: $$x=a\cos\theta$$

$$ A=4a^2\int_{0}^{\pi/2} \left(\sin^2\frac{\theta}{2} -2\sin^4\frac{\theta}{2}\right)d\theta $$

$$ A=\frac{\pi a^2}{2} $$

Answer: $$\boxed{\frac{\pi a^2}{2}}$$

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End of Chapter Section

Application of Double Integration

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 area bounded by $2x-3y+4=0$, $x+y-3=0$, $y=0$.

🔽 Show Step-by-Step Solution

Vertices: $$(-2,0), (3,0), (1,2)$$

$$A=\int_{0}^{2}\int_{\frac{3y-4}{2}}^{3-y} dx\,dy$$

$$A=\int_{0}^{2}\left(5-\frac{5y}{2}\right) dy$$

$$A=5$$

Final Answer: $$\boxed{5}$$

Example 4

Find the area bounded by $x^2=4(y+2)$ and $x^2=3-y$.

🔽 Show Step-by-Step Solution

Intersection: $$4(y+2)=3-y$$ $$y=-1,\quad x=\pm2$$

$$A=\int_{-2}^{2}\int_{\frac{x^2}{4}-2}^{3-x^2} dy\,dx$$

$$A=\int_{-2}^{2}\left(5-\frac{5x^2}{4}\right) dx$$

$$A=\frac{172}{15}$$

Final Answer: $$\boxed{\frac{172}{15}}$$

Example 5

Find the area bounded by $x(x^2+y^2)=a(x^2-y^2)$.

🔽 Show Step-by-Step Solution

Using polar substitution: $$r = a\cos(2\theta)\sec\theta$$

Area of full loop: $$A=4a^2\int_{0}^{\pi/2} \left(\sin^2\frac{\theta}{2} -2\sin^4\frac{\theta}{2}\right)d\theta$$

After evaluation: $$A=\frac{\pi a^2}{2}$$

Final Answer: $$\boxed{\frac{\pi a^2}{2}}$$

Conclusion

Double integrals are powerful tools for computing area, volume, and many physical applications.

LDLT Factorization

LDL^T Factorization – Detailed Numerical Examples

Dr. Brajesh Kumar Jha

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General Formula

For symmetric matrix $A$:

$$ A = L D L^T $$ $$ d_k = a_{kk} - \sum_{j=1}^{k-1} l_{kj}^2 d_j $$ $$ l_{ik} = \frac{1}{d_k} \left( a_{ik} - \sum_{j=1}^{k-1} l_{ij} l_{kj} d_j \right) $$

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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

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

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Example 2

$$ A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix} $$

Step 1

$$ d_1 = 2 $$ $$ l_{21} = -\frac{1}{2}, \quad l_{31} = \frac{1}{2} $$

Step 2

$$ d_2 = 2 - \left(-\frac{1}{2}\right)^2(2) $$ $$ = 2 - \frac{1}{2} = \frac{3}{2} $$ $$ l_{32} = \frac{1}{\frac{3}{2}} \left( -1 - \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)(2) \right) $$ $$ = \frac{2}{3}\left(-1+\frac{1}{2}\right) = -\frac{1}{3} $$

Step 3

$$ d_3 = 2 - \left(\frac{1}{2}\right)^2(2) - \left(-\frac{1}{3}\right)^2\left(\frac{3}{2}\right) $$ $$ = 2 - \frac{1}{2} - \frac{1}{6} = \frac{4}{3} $$

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Example 3

$$ A = \begin{bmatrix} 9 & 3 & 6 \\ 3 & 5 & 4 \\ 6 & 4 & 10 \end{bmatrix} $$

Step 1

$$ d_1 = 9 $$ $$ l_{21} = \frac{3}{9}=\frac{1}{3}, \quad l_{31} = \frac{6}{9}=\frac{2}{3} $$

Step 2

$$ d_2 = 5 - \left(\frac{1}{3}\right)^2(9) $$ $$ = 5 - 1 = 4 $$ $$ l_{32} = \frac{1}{4} \left( 4 - \frac{2}{3}\cdot\frac{1}{3}\cdot9 \right) = \frac{1}{2} $$

Step 3

$$ d_3 = 10 - \left(\frac{2}{3}\right)^2(9) - \left(\frac{1}{2}\right)^2(4) $$ $$ = 10 - 4 - 1 = 5 $$

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Indefinite Matrix Example

$$ A = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 1 & 2 \\ 0 & 2 & 1 \end{bmatrix} $$

Step 1

$$ d_1 = 1 $$ $$ l_{21} = 2, \quad l_{31}=0 $$

Step 2

$$ d_2 = 1 - (2)^2(1) $$ $$ = 1 - 4 = -3 $$ $$ l_{32} = \frac{1}{-3}(2) = -\frac{2}{3} $$

Step 3

$$ d_3 = 1 - \left(-\frac{2}{3}\right)^2(-3) $$ $$ = 1 + \frac{4}{3} = \frac{7}{3} $$

Since $d_2 < 0$, the matrix is indefinite.

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Comparison: LDL^T vs Cholesky

LDLT	Cholesky
$A = L D L^T$	$A = L L^T$
No square roots	Uses square roots
Works for indefinite matrices	Only positive definite

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Conclusion

• If all $d_i > 0$ → Positive definite matrix
• If some $d_i < 0$ → Indefinite matrix
• LDL^T avoids square roots

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: $$\iiint_V (x^2+y^2+z^2)\, dV$$ over sphere of radius $R$.

11. Find the volume of upper hemisphere $$x^2+y^2+z^2=16$$

Cylindrical Coordinates

12. Evaluate: $$\iiint_V r \, dV$$ where $x^2+y^2=4$, $0 \le z \le 3$.

13. Find volume enclosed by $$z=9-x^2-y^2$$ and $z=0$.

14. Evaluate: $$\iiint_V z\, dV$$ inside cylinder $x^2+y^2=1$, $0 \le z \le 2$.

Change of Variables

15. Evaluate using $u=x+y$, $v=x-y$: $$\iiint_V (x-y)\, dV$$

16. Evaluate: $$\iiint_V e^{-(x^2+y^2+z^2)} dV$$ over entire space.

Applications

17. Find centroid of solid $0 \le x \le a$, $0 \le y \le b$, $0 \le z \le c$.

18. Find moment of inertia about $z$-axis for cylinder $x^2+y^2 \le R^2$, $0 \le z \le h$.

19. Evaluate over tetrahedron $x=0$, $y=0$, $z=0$, $x+y+z=1$: $$\iiint_V xyz\, dV$$

20. Find volume common to sphere $$x^2+y^2+z^2=9$$ and cylinder $x^2+y^2=4$.

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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.

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#

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 of a solid
• Mass (if density is given)
• Center of mass
• Physical quantities in engineering

Imagine breaking a solid object into many very small cubes. If we add up the value of the function at each cube times its small volume, we get the triple integral.

If we integrate just $1$, we simply get the volume:

$$ \iiint_V 1\, dV = \text{Volume} $$

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3. Geometric Interpretation (3D Illustration)

Small Volume Elements Inside a Solid:

Projection of a Solid onto the xy-plane:

Common Coordinate Systems:

• Cartesian: $dV = dx\,dy\,dz$
• Cylindrical: $dV = r\,dr\,d\theta\,dz$
• Spherical: $dV = \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta$

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Summary

A triple integral allows us to measure quantities distributed throughout a three-dimensional region. It generalizes area (double integrals) to volume and physical applications in space.

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Triple Integrals – Solved Examples

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Example 1

Evaluate:

$$ \int_{0}^{1}\int_{0}^{1-y}\int_{0}^{1-y-z} z \, dx \, dz \, dy $$

Solution:

$$ = \int_{0}^{1}\int_{0}^{1-y} z(1-y-z)\, dz\, dy $$ $$ = \int_{0}^{1}\int_{0}^{1-y} \left[(1-y)z - z^2\right] dz\, dy $$ $$ = \int_{0}^{1} \left[ \frac{(1-y)z^2}{2} - \frac{z^3}{3} \right]_0^{1-y} dy $$ $$ = \int_{0}^{1} \left[ \frac{(1-y)^3}{2} - \frac{(1-y)^3}{3} \right] dy $$ $$ = \frac{1}{6}\int_{0}^{1}(1-y)^3 dy $$ $$ = \frac{1}{6}\left[\frac{(1-y)^4}{-4}\right]_0^1 $$ $$ = \frac{1}{24} $$

Geometric Region:

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Example 2

Evaluate:

$$ \iiint_V dV $$ where $$1 \le x \le 2,\quad 2 \le y \le 4,\quad 2 \le z \le 5.$$

Solution:

$$ = \int_{2}^{4}\int_{1}^{2}\int_{2}^{5} dz\, dx\, dy $$ $$ = \int_{2}^{4}\int_{1}^{2} 3\, dx\, dy $$ $$ = \int_{2}^{4} 3(1)\, dy $$ $$ = 3[y]_2^4 $$ $$ = 6 $$

Region (Rectangular Box):

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Example 3

Evaluate:

$$ \iiint_V 2x\, dV $$ where $V$ lies under the plane $$2x + 3y + z = 6$$ in the first octant.

Region Description:

$$ 0 \le z \le 6 - 2x - 3y $$ Projection on $xy$-plane: $$ 0 \le x \le 3, \quad 0 \le y \le \frac{6-2x}{3} $$

Integral:

$$ \int_0^3 \int_0^{(6-2x)/3} 2x(6-2x-3y)\, dy\, dx $$ $$ = \int_0^3 \left(\frac{4}{3}x^3 - 8x^2 + 12x\right) dx $$ $$ = 9 $$

Solid Figure:

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Example 4

Evaluate:

$$ \iiint_V z(x^2+y^2)\, dV $$ over the cylinder $$x^2+y^2 \le 1$$ between $z=2$ and $z=3$.

Integral Setup:

$$ = \int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \int_2^3 z(x^2+y^2)\, dz\, dx\, dy $$ $$ = \frac{5}{2}\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} (x^2+y^2)\, dx\, dy $$ Using $y=\sin\theta$, $$ = \frac{10}{3}\int_0^{\pi/2} \cos^4\theta\, d\theta + 10\int_0^{\pi/2} \sin^2\theta\cos^2\theta\, d\theta $$ $$ = \frac{5\pi}{4} $$

Cylinder Region:

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Final Answers

• Example 2: 1/24
• Example 3: 6
• Example 4: 9
• Example 5: 5π/4

Double Integration

Double Integration: Solved Examples with Graphs

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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 $$R = [0, 1] \times [0, 2]$$.

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 $$a$$.

$$\iint_D dA = \int_{0}^{2\pi} \int_{0}^{a} r \, dr \, d\theta$$ Step 1: Radial Integral $$\int_{0}^{a} r \, dr = \left[ \frac{r^2}{2} \right]_{0}^{a} = \frac{a^2}{2}$$ Step 2: Angular Integral $$\int_{0}^{2\pi} \frac{a^2}{2} \, d\theta = \left[ \frac{a^2 \theta}{2} \right]_{0}^{2\pi} = \pi a^2$$

Final Answer: $$\pi a^2$$

Example 3: Triangular Region

Evaluate \[ \iint_R xy\,dA \] where \(R\) is bounded by \(y=0,\; y=x,\; x=2\).

\[ \int_0^2\int_0^x xy\,dy\,dx = \int_0^2 \frac{x^3}{2}\,dx = 2 \] Answer: \( \boxed{2} \)

Example 4: Area Between Curves

Find the area bounded by \(y=x^2\) and \(y=4\).

\[ \text{Area} = \int_{-2}^{2}\int_{x^2}^{4} dy\,dx = \int_{-2}^{2} (4-x^2)\,dx = \frac{32}{3} \] Answer: \( \boxed{\frac{32}{3}} \)

Example 5: Circular Region

Evaluate \[ \iint_R (x^2+y^2)\,dA \] where \(x^2+y^2 \le 1\).

Using polar coordinates: \[ \int_0^{2\pi}\int_0^1 r^3\,dr\,d\theta = \frac{\pi}{2} \] Answer: \( \boxed{\frac{\pi}{2}} \)

Example 6: Region Between Two Curves

Evaluate \[ \iint_R y\,dA \] where \(y=x^2\) and \(y=x\).

\[ \int_0^1\int_{x^2}^{x} y\,dy\,dx = \int_0^1 \frac{x^2-x^4}{2}\,dx = \frac{1}{15} \] Answer: \( \boxed{\frac{1}{15}} \)

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Double Integration: 7 Solved Examples with Step-by-Step Solutions

Double integration is used to calculate the area of a 2D region or the volume under a 3D surface. Here are four essential examples.

Example 1: Rectangular Region

Evaluate $$\iint_R (8x + 6y) \, dA$$ where $$R = [0, 1] \times [0, 2]$$.

Figure 1: Rectangular Domain on the xy-plane

Step 1: Setup $$\int_{0}^{2} \int_{0}^{1} (8x + 6y) \, dx \, dy$$ Step 2: Inner Integration (dx) $$[4x^2 + 6xy]_{0}^{1} = 4 + 6y$$ Step 3: Outer Integration (dy) $$\int_{0}^{2} (4 + 6y) \, dy = [4y + 3y^2]_{0}^{2} = 8 + 12 = 20$$

Result: 20

Example 8: Triangular Region (Type I)

Evaluate $$\iint_D xy^2 \, dA$$ for the triangle with vertices (0,0), (2,0), (2,1).

Figure 2: Triangle bounded by y = 0, x = 2, and y = x/2

Step 1: Setup $$\int_{0}^{2} \int_{0}^{x/2} xy^2 \, dy \, dx$$ Step 2: Inner Integration (dy) $$[x \frac{y^3}{3}]_{0}^{x/2} = \frac{x}{3} \cdot \frac{x^3}{8} = \frac{x^4}{24}$$ Step 3: Outer Integration (dx) $$\int_{0}^{2} \frac{x^4}{24} \, dx = [\frac{x^5}{120}]_{0}^{2} = \frac{32}{120} = 4/15$$

Result: 4/15

Example 9: Polar Coordinates

Evaluate $$\iint_D (x^2 + y^2) \, dA$$ where $$D$$ is the unit circle.

Figure 3: Unit Circle in Polar Form ($r=1, \theta=2\pi$)

Step 1: Setup (Polar) $$\int_{0}^{2\pi} \int_{0}^{1} (r^2)r \, dr \, d\theta$$ Step 2: Solve $$\int_{0}^{2\pi} [\frac{r^4}{4}]_0^1 \, d\theta = \int_{0}^{2\pi} \frac{1}{4} d\theta = \frac{2\pi}{4} = \pi/2$$

Result: π/2

Example 10: Volume of Paraboloid

Find volume under $$z = 1 - x^2 - y^2$$ above the xy-plane.

Figure 4: 3D Visualization of the Paraboloid

Solution: Using polar conversion $$1-(x^2+y^2) = 1-r^2$$: $$\int_{0}^{2\pi} \int_{0}^{1} (1-r^2)r \, dr \, d\theta = \pi/2$$

Result: π/2

Graphs generated for educational purposes. For interactive 3D plots, use GeoGebra 3D.

Live Calculator

Quick Formulas

Polar $$dA$$: $$r \, dr \, d\theta$$

Volume: $$\iint_D f(x,y) \, dA$$

Average Value: $$\frac{1}{Area(D)} \iint_D f \, dA$$

Mass: $$\iint_D \rho(x,y) \, dA$$

Weighted Sampling

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)

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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 $[1,3,6] \Rightarrow [0.1,0.3,0.6]$

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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.

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Four Examples of Weighted Sampling

• Lottery: More tickets $\Rightarrow$ higher winning chance
• Online advertising: Higher bids shown more often
• Survey sampling: Oversampling rare groups
• Game loot drops: Common items have higher drop rates

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Weighted Sampling in Machine Learning

• Class imbalance: Minority classes sampled more frequently
• Mini-batch construction: Hard examples sampled more often
• Reinforcement learning: Prioritized experience replay
• Monte Carlo methods: Importance sampling

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Types of Weighted Sampling

1. With replacement
   Probabilities remain constant across draws
2. Without replacement
   Probabilities change after each selection
3. Stratified weighted sampling
   Sampling within predefined groups
4. Importance sampling
   Sampling with probability correction

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Importance Sampling

Goal: estimate an expectation under distribution $p(x)$ while sampling from $q(x)$.

$ \mathbb{E}_{p}[f(x)] = \mathbb{E}_{q}\left[f(x)\frac{p(x)}{q(x)}\right] $

Importance weight:

$ w(x) = \frac{p(x)}{q(x)} $

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Weighted Sampling vs Loss Weighting

Aspect	Weighted Sampling	Loss Weighting
Data frequency	Changes	Unchanged
Loss magnitude	Unchanged	Scaled

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Loss Weighting Formula

Instead of changing how often samples appear, loss weighting scales their contribution to the loss:

$ \mathcal{L} = \sum_i w_i \cdot \ell(y_i, \hat{y}_i) $

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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 → higher chance of being selected
Lower weight → lower chance (but usually not zero)

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Mathematical Definition

Given n items with weights:

w₁, w₂, ..., w_n

The probability of selecting item i is:

P(i) = w_i / Σ_j=1ⁿ w_j

Example: weights [1, 3, 6] → probabilities [0.1, 0.3, 0.6]

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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.

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Four Examples of Weighted Sampling

• Lottery: More tickets mean higher winning chances
• Online advertising: Higher bids are shown more often
• Survey sampling: Rare groups are oversampled
• Game loot drops: Common items have higher drop rates

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Weighted Sampling in Machine Learning

• Class imbalance: Minority classes sampled more frequently
• Mini-batch construction: Hard examples sampled more often
• Reinforcement learning: Prioritized experience replay
• Monte Carlo methods: Importance sampling

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Types of Weighted Sampling

1. With replacement
   Items can be selected multiple times (e.g., bootstrapping)
2. Without replacement
   Items are removed after selection (e.g., surveys)
3. Stratified weighted sampling
   Sampling within predefined groups (strata)
4. Importance sampling
   Sampling from one distribution and correcting with weights

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Importance Sampling Formula

Goal: estimate an expectation under distribution p(x) while sampling from q(x).

E_p[f(x)] = E_q[ f(x) · p(x) / q(x) ]

Weight:

w(x) = p(x) / q(x)

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Weighted Sampling vs Loss Weighting

Aspect	Weighted Sampling	Loss Weighting
Changes data frequency	Yes	No
Changes loss magnitude	No	Yes
Affects mini-batches	Yes	No

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Loss Weighting Formula

Instead of changing how often samples appear, loss weighting scales their contribution to the loss:

L = Σ w_i · ℓ(y_i, ŷ_i)

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Key Takeaway

Weighted sampling controls what the model sees more often.
Loss weighting controls how much mistakes matter.

Key Takeaway

Weighted sampling controls what the model sees more often.
Loss weighting controls how much mistakes matter.