Module 3: Double and Iterated Integrals PDF

# Module 3: Double and Iterated Integrals over Rectangles ## Definition and Formula - **The Volume of a Solid:** The volume of a solid under the surface *z = f(x, y)* and above a rectangular region *R* can be found using a double integral. - **Double integral formula:** ``` Volume = lim Sn...

# Module 3: Double and Iterated Integrals over Rectangles ## Definition and Formula - **The Volume of a Solid:** The volume of a solid under the surface *z = f(x, y)* and above a rectangular region *R* can be found using a double integral. - **Double integral formula:** ``` Volume = lim Sn = ∫∫_R f(x, y) dA, where n approaches infinity. ``` ## Fubini's Theorem - **Fubini's Theorem (First form):** - **If f(x, y) is continuous throughout the rectangular region R: a ≤ x ≤ b, c ≤ y ≤ d**, then: ``` ∫∫_R F(x, y) dA = ∫_c^d ∫_a^b f(x, y) dx dy = ∫_a^b ∫_c^d f(x, y) dy dx. ``` - **Fubini's Theorem (Stronger form)**: - **If R is defined by a ≤ x ≤ b, g₁(x) ≤ y ≤ g2(x), with g₁ and g2 continuous on [a, b]**: ``` ∫∫_R f(x, y) dA = ∫_a^b ∫_g₁(x)^g₂(x) f(x, y) dy dx. ``` - **If R is defined by c ≤ y ≤ d, h₁(y) ≤ x ≤ h₂(y), with h₁ and h₂ continuous on [c, d]**: ``` ∫∫_R f(x, y) dA = ∫_c^d ∫_h₁(y)^h₂(y) f(x, y) dx dy. ``` ## Examples 1. **Finding the Volume of a Prism:** - **Problem:** Find the volume of the prism whose base is the triangle in the xy-plane bounded by the x-axis and the lines y = x and x = 1 and whose top lies in the plane z = f(x, y) = 3 - x - y. - **Solution:** The volume of the prism is: ``` ∫∫_R (3 - x - y) dA = ∫_0^1 ∫_0^x (3 - x - y) dy dx = 1 ``` 2. **Finding the Volume of a Wedge-like Solid:** - **Problem:** Find the volume of the wedgelike solid that lies beneath the surface z = 16 - x² - y² and above the region R bounded by the curve y = 2√x, the line y = 4x - 2, and the x-axis. - **Solution:** The volume of the wedgelike solid is: ``` ∫∫_R (16 - x² - y²) dA = ∫_0^1 ∫_2√x^(4x - 2) (16 - x² - y²) dy dx = 36/5 ``` 3. **Finding the Volume of a Solid Bounded by a Paraboloid and a Triangle:** - **Problem:** Find the volume of the solid that is bounded above by the paraboloid z = x² + y² and below by the triangle enclosed by the lines y = x, x = 0, and x + y = 2 in the xy-plane. - **Solution:** The volume of the solid is: ``` ∫∫_R (x² + y²) dA = ∫_0^1 ∫_x^(2 - x) (x² + y²) dy dx = 1/2 ``` 4. **Finding the Volume of a Solid Bounded by a Cylinder and a Parabola:** - **Problem:** Find the volume of the solid that is bounded above by the cylinder z = x² and below by the region enclosed by the parabola y = 2 x² and the line y = x in the xy-plane. - **Solution:** The volume of the solid is: ``` ∫∫_R x² dA = ∫_0^1 ∫_2x²^x x² dy dx = 1/15 ``` 5. **Finding the Volume of a Solid Bounded by a Plane and a Parabola:** - **Problem:** Find the volume of the solid whose base is the region in the xy-plane that is bounded by the parabola y = 4 x² and the line y = 3x, while the top of the solid is bounded by the plane z = x + 4. - **Solution:** The volume of the solid is: ``` ∫∫_R (x + 4) dA = ∫_0^(4/3) ∫_4x²^(3x) (x + 4) dy dx = 32/9 ``` 6. **Finding the Volume of a Solid Bounded by a Cylinder and a Plane:** - **Problem:** Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder x² + y² = 4, and the plane z + y = 3. - **Solution:** The volume of the solid is: ``` ∫∫_R (3 - y) dA = ∫_0^2 ∫_0^√(4 - y²) (3 - y) dx dy = 8π/3 ``` ## Applications 1. **Area of a Region:** The area of a closed, bounded plane region *R* is given by: ``` A = ∫∫_R dA ``` 2. **Average Value of a Function:** The average value of *f(x, y)* over region *R* is: ``` Average value of f over R = (1/area of R) ∫∫_R f dA. ``` 3. **Population Density:** The total population of a region with a given population density *f(x, y)* can be found by integrating the function over the region. - For example, if the population density of a region is given by *f(x, y) = 20,000e^(-x²-y²)*, the total population in the region bounded by *x² + y² = 1* can be calculated using a double integral. 4. **Electric Charge:** The total electric charge *Q* distributed over a region *D* with charge density *σ(x, y)* is given by: ``` Q = ∫∫_D σ(x, y) dA. ``` ## Change of Variables - **Substitution for Double Integrals:** If a transformation *x = g(u, v)*, *y = h(u, v)* is one-to-one on the interior of G, then: ``` ∫∫_R f(x, y) dx dy = ∫∫_G |J(u, v)| f(g(u, v), h(u, v)) du dv, ``` - **Where J(u, v)| is the Jacobian of the transformation:** ``` J(u, v) = |(∂x/∂u)(∂y/∂v) - (∂x/∂v)(∂y/∂u)| ``` ## Polar Coordinates - **Change of Variables in Polar Coordinates:** ``` ∫∫_R f(x, y) dx dy = ∫∫_G f(r cos θ, r sin θ) r dr dθ, ``` ## Examples in Polar Coordinates 1. **Finding the Area Inside a Cardioid and Outside a Circle:** - **Problem:** Find the limits of integration for integrating f(r, θ) over the region R that lies inside the cardioid r = 1 + cos θ and outside the circle r = 1. - **Solution:** ``` ∫∫_R f(r, θ) dA = ∫_0^(2π) ∫_1^(1 + cos θ) f(r, θ) r dr dθ ``` 2. **Evaluating a Double Integral in Polar Coordinates:** - **Problem:** Evaluate ∫∫ r³ dr dθ over the area bounded between the circles r = 2 cos θ and r = 4 cos θ. - **Solution:** ``` ∫∫_R r³ dr dθ = ∫_0^(π/2) ∫_2 cos θ^(4 cos θ ) r³ dr dθ = 63π/16 ``` 3. **Finding the Area of a Region in Polar Coordinates:** - **Problem:** Find the area enclosed by the lemniscate r² = 4 cos 2θ. - **Solution:** ``` A = ∫∫_R r dr dθ = ∫_0^(π/4) ∫_0^√(4 cos 2θ) r dr dθ = 4 ``` 4. **Finding the Area Inside a Circle and Outside a Cardioid:** - **Problem:** Find the area of the region that lies inside the circle r = 3 sin θ and outside the cardioid r = 1 + sin θ. - **Solution:** ``` ∫∫_R dA = ∫_0^(π) ∫_1+sinθ^(3 sin θ) r dr dθ = π - 1/4 ``` ## Moments and Centroids - **Mass:** The mass *M* of a thin plate covering a region *R* with density *δ(x, y)* is given by: ``` M = ∫∫_R δ(x, y) dA. ``` - **First Moments:** - About the *x*-axis: ``` Mx = ∫∫_R y δ(x, y) dA. ``` - About the *y*-axis: ``` My = ∫∫_R x δ(x, y) dA. ``` - **Center of Mass:** ``` x = My/M, y = Mx/M. ``` - **Moments of Inertia (Second Moments):** - About the *x*-axis: ``` Ix = ∫∫_R y² δ(x, y) dA. ``` - About the *y*-axis: ``` Iy = ∫∫_R x² δ(x, y) dA. ``` - About an axis *L:* ``` IL = ∫∫_R r²(x, y) δ(x, y) dA, where r(x, y) = distance from (x, y) to L. ``` - About the origin (polar moment): ``` Io = ∫∫_R (x² + y²) δ(x, y) dA. ``` - **Radii of Gyration:** - About the *x*-axis: ``` Rx = √(Ix/M). ``` - About the *y*-axis: ``` Ry = √(Iy/M). ``` - About the origin: ``` Ro = √(Io/M). ``` ## Centroid - **Centroid of a Figure with Constant Density:** If the density of a geometric figure is constant, the centroid is the same as its center of mass. We can set the density to 1 and calculate the center of mass using the formulas above. - **Finding the Centroid of Geometric Figures:** - **Find the mass:** Calculate the mass *M* of the figure using the double integral. - **Find the first moments:** Calculate the first moments *Mx* and *My* of the figure. - **Calculate the centroid:** Calculate the coordinates of the centroid (x, y) using the formulas: *x = My/M* and *y = Mx/M*. ## Examples of Moments and Centroids 1. **Finding the Mass, Moments, and Center of Mass of a Triangular Lamina:** - **Problem:** Find the mass and center of mass of a triangular lamina with vertices (0, 0), (1, 0), and (0, 2) if the density function is p(x, y) = 1 + 3x + y. - **Solution:** - **Mass:** ``` M = ∫∫_R p(x, y)dA = ∫_0^1 ∫_0^(2-2x) (1 + 3x + y) dy dx = 5/3 ``` - **First Moments:** ``` Mx = ∫∫_R y p(x, y)dA = ∫_0^1 ∫_0^(2-2x) y(1 + 3x + y) dy dx = 1/3 My = ∫∫_R x p(x, y)dA = ∫_0^1 ∫_0^(2-2x) x(1 + 3x + y) dy dx = 1/2 ``` - **Center of Mass:** ``` x = My/M = (1/2) / (5/3) = 3/10 y = Mx/M = (1/3) / (5/3) = 1/5 ``` Therefore, the center of mass is at (3/10, 1/5). 2. **Finding the Mass and Moment of Inertia of a Thin Plate:** - **Problem:** Find the mass and moment of inertia about the *x*-axis of a thin plate bounded by the parabola *x = y - y²* and the line *x + y = 0* if *δ(x, y) = x + y*. - **Solution:** - **Mass:** ``` M = ∫∫_R (x + y)dA = ∫_0^1 ∫_(y-y²)^(-y) (x + y)dx dy = 1/3 ``` - **Moment of Inertia about the *x*-axis:** ``` Ix = ∫∫_R y² (x + y)dA = ∫_0^1 ∫_(y-y²)^(-y) y² (x + y) dx dy = 1/10 ``` - **Radius of Gyration about the *x*-axis:** ``` Rx = √(Ix/M) = √(1/10 / 1/3) = √(3/10) ``` 3. **Finding the Center of Mass and Moment of Inertia of a Thin Plate:** - **Problem:** Find the center of mass and moment of inertia about the *x*-axis of a thin plate bounded by the curves *x = y²* and *x = 2y - y²* if the density at the point *(x, y)* is *δ(x, y) = y + 1*. - **Solution:** - **Mass:** ``` M = ∫∫_R (y + 1) dA = ∫_0^1 ∫_y²^(2y-y²) (y + 1) dx dy = 1/3 ``` - **Moment of Inertia about *x*-axis:** ``` Ix = ∫∫_R y² (y + 1) dA = ∫_0^1 ∫_y²^(2y-y²) y² (y + 1) dx dy = 11/60 ``` - **Radius of Gyration about the *x*-axis:** ``` Rx = √(11/60 / 1/3) = √11/20 ``` ## Change of Variables with Multiple Integrals 1. **Type 1 and Type 2 Regions:** Areas and volumes can be calculated using multiple integrals. - **Type 1 regions:** The region is bounded by the curves *y* = *g₁*(x) and *y* = *g₂*(x)* between the limits of *a* ≤ *x* ≤ *b*. - **Type 2 regions:** The region is bounded by the curves *x* = *h₁*(y) and *x* = *h₂*(y)* between the limits of *c* ≤ *y* ≤ *d*. 2. **Changing the Order of Integration:** The order of integration can sometimes be switched to make the problem easier to solve. * **Example:** Let's say we want to evaluate the double integral: ``` ∫_0^1 ∫_0^√(1 - x²) (x² + y²) dy dx ``` - Firstly, identify the limits of integration for the inner and outer integrals: - The inner integral has limits from *y = 0* to *y = √(1 - x²)* - The outer integral has limits from *x = 0* to *x = 1*. - Next, identify the type of region that these limits represent: This is a Type 1 region where *y* = *g₁*(x)* and *y* = *g₂*(x)* - Finally, swap the order of integration by changing the limits: ``` ∫_0^1 ∫_0^√(1 - y²) (x² + y²) dx dy ``` - Now, the limits for the inner integral are *x = 0* to *x = √(1 - y²)*, and the limits for the outer integral are *y = 0* to *y = 1*. We have switched to a Type 2 region. ## Change of Variables with Polar Coordinates - **Polar coordinates:** A point in the *xy*-plane can be represented using polar coordinates *r* and *θ*. - **Conversion Formulas:** - *x = r cos θ* - *y = r sin θ* - ** Jacobian Formula:** ``` |J(r, θ)| = |(∂x/∂r)(∂y/∂θ) - (∂x/∂θ)(∂y/∂r)| = r ``` - **Evaluating Double Integrals in Polar Coordinates:** Replacing *x* and *y* with their corresponding polar coordinates, and the *dA* term with *r dr dθ*, allows you to evaluate double integrals using polar coordinates. ## Examples of Change of Variables with Polar Coordinates 2. **Evaluating an Integral in Polar Coordinates:** - **Problem:** Evaluate ∫∫_R (x² + y²) dA, where *R* is the semicircular region bounded by the *x*-axis and the curve *y = √(1 - x²)*. - **Solution:** - **Convert the limits:** - *x = -√(1 - y²)* to *x = √(1 - y²)* becomes *r = 0* to *r = 1*. - From *y = 0* to *y = 1* becomes *θ = 0* to *θ = π*. ``` ∫∫_R (x² + y²) dA = ∫_0^π ∫_0^1 (r² cos^2 θ + r² sin^2 θ) r dr dθ = ∫_0^π ∫_0^1 r³ dr dθ = ∫_0^π (1/4) dθ = π/4. ``` 4. **Finding the Area Enclosed by a Cardioid:** - **Problem:** Find the area enclosed by one loop of the four-leaved rose *r = cos 2θ*. - **Solution:** - Since one loop is generated over the interval 0 ≤ θ ≤ π/4, the area is represented by: ``` ∫_0^(π/4) ∫_0^(cos 2θ) r dr dθ = ∫_0^(π/4) (1/2) cos² 2θ dθ = π/8 ``` 5. **Finding the Area Inside a Circle and Outside a Cardioid:** - **Problem:** Find the area of the region that lies inside the circle *r = 3 sin θ* and outside the cardioid *r = 1 + sin θ*. - **Solution:** - **Find the intersection points:** - We can equate the equations to find the intersection points: ``` 1 + sin θ = 3 sin θ 2 sin θ = 1 sin θ = 1/2 θ = π/6, 5π/6 ``` - **Set up the limits:** - *r = 1 + sin θ* from *θ = π/6* to *θ = 5π/6* (for the cardioid). - *r = 3 sin *θ* from *θ = 0* to *θ = π* ( for the circle). ``` ∫∫_R dA = ∫_π/6^(5π/6) ∫_1+sin θ^(3 sin θ) r dr dθ + ∫_0^(π/6) ∫_0^(3 sin θ) r dr dθ + ∫_(5π/6)^π ∫_0^(3 sin θ) r dr dθ ``` - Simplifying by using the symmetry of the region, we obtain: ``` ∫∫_R dA = 2 ∫_π/6^(π/2) ∫_1+sin θ^(3 sin θ) r dr dθ + 2 ∫_0^(π/6) ∫_0^(3 sin θ) r dr dθ = π - 1/4 ```