All Study Guides Calculus IV Unit 12
∞ Calculus IV Unit 12 – Applications of Double IntegralsDouble integrals extend single integrals to functions of two variables, allowing us to calculate areas, volumes, and other properties of complex shapes. They're evaluated using iterated integrals, where we integrate with respect to one variable at a time.
Applications of double integrals are vast, from finding centers of mass to calculating moments of inertia. They're used in physics, engineering, and other fields to solve real-world problems involving variable densities, electric charges, and gravitational potentials.
Key Concepts and Definitions
Double integrals extend the concept of single integrals to functions of two variables
Iterated integrals evaluate double integrals by integrating with respect to one variable at a time
Compute the inner integral first, treating the other variable as a constant
Then evaluate the outer integral using the result from the inner integral
Fubini's Theorem states that if f ( x , y ) f(x,y) f ( x , y ) is continuous over a closed, bounded region R R R , the double integral of f f f over R R R equals the iterated integral
The order of integration can be interchanged if the function is continuous and the region is bounded
Jacobian determinant is used when changing variables in double integrals
For polar coordinates, the Jacobian is r r r , so d A = r d r d θ dA = r \, dr \, d\theta d A = r d r d θ
Double integrals can be used to find volumes, areas, centers of mass, and moments of inertia
Double Integrals in Rectangular Coordinates
Double integrals in rectangular coordinates are written as ∬ R f ( x , y ) d A \iint_{R} f(x,y) \, dA ∬ R f ( x , y ) d A
The region R R R is typically described by the bounds of x x x and y y y
Evaluate the double integral by iterating the integrals with respect to x x x and y y y
Choose the order of integration based on the region's description
Sketch the region to determine the appropriate bounds for each integral
Example: ∫ 0 1 ∫ 0 1 − x x y d y d x \int_{0}^{1} \int_{0}^{1-x} xy \, dy \, dx ∫ 0 1 ∫ 0 1 − x x y d y d x integrates x y xy x y over the triangular region bounded by y = 0 y=0 y = 0 , x = 0 x=0 x = 0 , and x + y = 1 x+y=1 x + y = 1
To change the order of integration, adjust the bounds accordingly
For the example above, changing to d y d x dy \, dx d y d x yields ∫ 0 1 ∫ 0 1 − y x y d x d y \int_{0}^{1} \int_{0}^{1-y} xy \, dx \, dy ∫ 0 1 ∫ 0 1 − y x y d x d y
Double Integrals in Polar Coordinates
Double integrals in polar coordinates are written as ∬ R f ( r , θ ) r d r d θ \iint_{R} f(r,\theta) \, r \, dr \, d\theta ∬ R f ( r , θ ) r d r d θ
The region R R R is typically described by the bounds of r r r and θ \theta θ
Convert the function f ( x , y ) f(x,y) f ( x , y ) to polar form f ( r , θ ) f(r,\theta) f ( r , θ ) using x = r cos θ x = r\cos\theta x = r cos θ and y = r sin θ y = r\sin\theta y = r sin θ
Determine the bounds for r r r and θ \theta θ based on the region's description in polar coordinates
Sketch the region to identify the appropriate bounds
Evaluate the double integral by iterating the integrals with respect to r r r and θ \theta θ
Example: ∫ 0 π / 2 ∫ 0 1 r 2 d r d θ \int_{0}^{\pi/2} \int_{0}^{1} r^2 \, dr \, d\theta ∫ 0 π /2 ∫ 0 1 r 2 d r d θ integrates r 2 r^2 r 2 over the quarter circle of radius 1 in the first quadrant
Polar coordinates simplify double integrals for regions with circular symmetry
Applications in Area Calculation
Double integrals can be used to calculate areas of regions in the plane
For a region R R R , the area is given by ∬ R 1 d A \iint_{R} 1 \, dA ∬ R 1 d A
In rectangular coordinates, ∬ R 1 d A = ∫ a b ∫ g 1 ( x ) g 2 ( x ) 1 d y d x \iint_{R} 1 \, dA = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} 1 \, dy \, dx ∬ R 1 d A = ∫ a b ∫ g 1 ( x ) g 2 ( x ) 1 d y d x
g 1 ( x ) g_1(x) g 1 ( x ) and g 2 ( x ) g_2(x) g 2 ( x ) are the lower and upper bounds of y y y in terms of x x x
In polar coordinates, ∬ R 1 d A = ∫ α β ∫ h 1 ( θ ) h 2 ( θ ) r d r d θ \iint_{R} 1 \, dA = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} r \, dr \, d\theta ∬ R 1 d A = ∫ α β ∫ h 1 ( θ ) h 2 ( θ ) r d r d θ
h 1 ( θ ) h_1(\theta) h 1 ( θ ) and h 2 ( θ ) h_2(\theta) h 2 ( θ ) are the lower and upper bounds of r r r in terms of θ \theta θ
Example: Find the area enclosed by the circle x 2 + y 2 = 4 x^2 + y^2 = 4 x 2 + y 2 = 4
In polar coordinates, the equation becomes r 2 = 4 r^2 = 4 r 2 = 4 , so r = 2 r = 2 r = 2
∫ 0 2 π ∫ 0 2 r d r d θ = 4 π \int_{0}^{2\pi} \int_{0}^{2} r \, dr \, d\theta = 4\pi ∫ 0 2 π ∫ 0 2 r d r d θ = 4 π
Volume Calculation Using Double Integrals
Double integrals can calculate volumes of solids by integrating cross-sectional areas
For a solid bounded by z = f ( x , y ) z = f(x,y) z = f ( x , y ) and the region R R R in the x y xy x y -plane, the volume is ∬ R f ( x , y ) d A \iint_{R} f(x,y) \, dA ∬ R f ( x , y ) d A
In rectangular coordinates, ∬ R f ( x , y ) d A = ∫ a b ∫ c d f ( x , y ) d y d x \iint_{R} f(x,y) \, dA = \int_{a}^{b} \int_{c}^{d} f(x,y) \, dy \, dx ∬ R f ( x , y ) d A = ∫ a b ∫ c d f ( x , y ) d y d x
Integrate the cross-sectional area f ( x , y ) f(x,y) f ( x , y ) over the region R R R
In polar coordinates, ∬ R f ( r , θ ) d A = ∫ α β ∫ g 1 ( θ ) g 2 ( θ ) f ( r , θ ) r d r d θ \iint_{R} f(r,\theta) \, dA = \int_{\alpha}^{\beta} \int_{g_1(\theta)}^{g_2(\theta)} f(r,\theta) \, r \, dr \, d\theta ∬ R f ( r , θ ) d A = ∫ α β ∫ g 1 ( θ ) g 2 ( θ ) f ( r , θ ) r d r d θ
Convert f ( x , y ) f(x,y) f ( x , y ) to polar form f ( r , θ ) f(r,\theta) f ( r , θ ) and integrate over the region R R R
Example: Find the volume of the solid bounded by z = x 2 + y 2 z = x^2 + y^2 z = x 2 + y 2 and z = 4 z = 4 z = 4
In rectangular coordinates, ∫ − 2 2 ∫ − 4 − x 2 4 − x 2 ( 4 − x 2 − y 2 ) d y d x \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} (4 - x^2 - y^2) \, dy \, dx ∫ − 2 2 ∫ − 4 − x 2 4 − x 2 ( 4 − x 2 − y 2 ) d y d x
Center of Mass and Moments of Inertia
Double integrals can find the center of mass and moments of inertia for thin plates
For a thin plate with density δ ( x , y ) \delta(x,y) δ ( x , y ) over a region R R R , the center of mass is ( x ˉ , y ˉ ) (\bar{x}, \bar{y}) ( x ˉ , y ˉ ) :
x ˉ = ∬ R x δ ( x , y ) d A ∬ R δ ( x , y ) d A \bar{x} = \frac{\iint_{R} x \, \delta(x,y) \, dA}{\iint_{R} \delta(x,y) \, dA} x ˉ = ∬ R δ ( x , y ) d A ∬ R x δ ( x , y ) d A and y ˉ = ∬ R y δ ( x , y ) d A ∬ R δ ( x , y ) d A \bar{y} = \frac{\iint_{R} y \, \delta(x,y) \, dA}{\iint_{R} \delta(x,y) \, dA} y ˉ = ∬ R δ ( x , y ) d A ∬ R y δ ( x , y ) d A
Moments of inertia measure a plate's resistance to rotational acceleration
I x = ∬ R y 2 δ ( x , y ) d A I_x = \iint_{R} y^2 \, \delta(x,y) \, dA I x = ∬ R y 2 δ ( x , y ) d A and I y = ∬ R x 2 δ ( x , y ) d A I_y = \iint_{R} x^2 \, \delta(x,y) \, dA I y = ∬ R x 2 δ ( x , y ) d A
For a plate with constant density δ \delta δ , the formulas simplify to:
x ˉ = ∬ R x d A ∬ R d A \bar{x} = \frac{\iint_{R} x \, dA}{\iint_{R} dA} x ˉ = ∬ R d A ∬ R x d A , y ˉ = ∬ R y d A ∬ R d A \bar{y} = \frac{\iint_{R} y \, dA}{\iint_{R} dA} y ˉ = ∬ R d A ∬ R y d A , I x = δ ∬ R y 2 d A I_x = \delta \iint_{R} y^2 \, dA I x = δ ∬ R y 2 d A , I y = δ ∬ R x 2 d A I_y = \delta \iint_{R} x^2 \, dA I y = δ ∬ R x 2 d A
Real-World Applications and Examples
Double integrals have numerous applications in physics, engineering, and other fields
Calculating the mass of a thin plate with variable density δ ( x , y ) \delta(x,y) δ ( x , y )
Mass = ∬ R δ ( x , y ) d A = \iint_{R} \delta(x,y) \, dA = ∬ R δ ( x , y ) d A
Finding the average value of a function f ( x , y ) f(x,y) f ( x , y ) over a region R R R
Average value = ∬ R f ( x , y ) d A ∬ R d A = \frac{\iint_{R} f(x,y) \, dA}{\iint_{R} dA} = ∬ R d A ∬ R f ( x , y ) d A
Determining the electric charge on a plate with charge density ρ ( x , y ) \rho(x,y) ρ ( x , y )
Total charge = ∬ R ρ ( x , y ) d A = \iint_{R} \rho(x,y) \, dA = ∬ R ρ ( x , y ) d A
Calculating the gravitational or electrostatic potential at a point due to a thin plate
Potential ∝ ∬ R δ ( x , y ) r d A \propto \iint_{R} \frac{\delta(x,y)}{r} \, dA ∝ ∬ R r δ ( x , y ) d A , where r r r is the distance from the point to ( x , y ) (x,y) ( x , y )
Example: Find the mass of a circular plate of radius 2 with density δ ( x , y ) = x 2 + y 2 \delta(x,y) = x^2 + y^2 δ ( x , y ) = x 2 + y 2
In polar coordinates, ∫ 0 2 π ∫ 0 2 r 2 ⋅ r d r d θ = 16 π 3 \int_{0}^{2\pi} \int_{0}^{2} r^2 \cdot r \, dr \, d\theta = \frac{16\pi}{3} ∫ 0 2 π ∫ 0 2 r 2 ⋅ r d r d θ = 3 16 π
Common Challenges and Problem-Solving Strategies
Setting up the correct bounds of integration based on the region's description
Sketch the region to visualize the bounds
Write the bounds in terms of the integration variables
Choosing the appropriate coordinate system (rectangular or polar) for the problem
Polar coordinates are often easier for regions with circular symmetry
Rectangular coordinates are better suited for regions bounded by straight lines
Evaluating integrals involving trigonometric or exponential functions
Use trigonometric identities and integration techniques (substitution, integration by parts)
Remembering to include the Jacobian determinant when changing variables
In polar coordinates, multiply the integrand by r r r to account for the Jacobian
Verifying that the integrand and region satisfy the conditions of Fubini's Theorem
Ensure the function is continuous over the closed, bounded region
Applying double integrals to real-world problems
Identify the physical quantity to be calculated (area, volume, mass, etc.)
Set up the integral with the appropriate integrand and region