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
2. Area of a Plane Region
The area of region $R$:
$$ A = \iint_R 1\, dA $$
Example Region (Circle)
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
5. Moments of Inertia
$$ I_x = \iint_R y^2 \rho(x,y)\, dA $$ $$ I_y = \iint_R x^2 \rho(x,y)\, dA $$ $$ I_0 = \iint_R (x^2 + y^2)\rho(x,y)\, dA $$
6. Probability Applications
If $f(x,y)$ is a joint probability density function:
$$ P((x,y)\in R) = \iint_R f(x,y)\, dA $$ $$ \iint_{\mathbb{R}^2} f(x,y)\, dA = 1 $$
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
End of Notes
Applications of Double Integrals
Area Enclosed by Plane Curves
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: $$\boxed{\frac{\pi a^2}{4}}$$
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: $$\boxed{18}$$
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: $$\boxed{5}$$
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: $$\boxed{\frac{172}{15}}$$
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: $$\boxed{\frac{\pi a^2}{2}}$$
End of Chapter Section
No comments:
Post a Comment