Powerpoint on double integrals

1. Double Integrals Jason Hsiao Roy Park Ben Lo

2. Double Integral and Fibuni’s Theorem The integral of an integral Another Method for finding Volume Mass density Centers of mass Joint probability Expected value Fibuni’s Theorem states that if f is continuous on a plane region R

3. Properties of Double Integrals The two intervals determine the region of integration R on the iterated integral

4. Example Problem **Do Inner Integral first! Integrate with respect to x. Treat y as a constant Integrate with respect to y NOTE: similar to partial derivatives Concerning treatment of variables as a constant

5. Example problem 2 ∫ ∫ (x2-2y2+1) dxdy ∫ [(x3)/3-2y2x+x] | dy ∫ [((64/3)-8y2+4)-(0 -0 +0)] dy [(64y)/3- (8y3)/3+4y] | [(64(2))/3-(8(23)/3+4(2)]-[(64(1))/3-(8(13))/3+4(1)] (128-64)/3+(-64+8)/3 +(8-4) 64/3-56/3+4 8/3+4 20/3 2 1 4 0 Integrate with respect to x. Treat y as constant 2 1 4 0 2 1 Integrate with respect to y 2 1

6. In mathematics, a planar laminais a closed surface of mass m and surface density such that:, over the closed surface. Planar laminas can be used to compute mass, electric charge, moments of inertia, or center of mass. Real Life Application

7. Suppose the lamina occupies a region D of the xy-plane and its density at a point (x,y) in D is given by ρ(x,y) where ρ is a continuous function on D. This means that: Ρ(x,y)=lim where ∆m and ∆A are the mass and area of a small rectangle that contains (x,y) and the limit is taken as the dimensions of the rectangle approach 0. Therefore we arrive at the definition of total mass in the lamina. All one has to do is find the double integral of the density function. m=∬ρ(x,y)dA Density and Mass ∆m ___ ∆A

8. Moments of Center of Mass The center of mass of a lamina with density function ρ(x,y) that occupies a region D. To find the center of mass we first have to find the moment of a particle about an axis, which is defined as the product of its mass and its directed distance from the axis. The moment of the entire lamina about the x-axis: Mx=∬yρ(x,y)dA Similarly, the moment about the y-axis: My=∬xρ(x,y)dA You can define the center of mass (α,ŷ) so that mα=My and mŷ=Mx The physical significance is that the lamina behaves as if its entire mass is concentrated at its center of mass. Thus, the lamina balances horizontally when supported at its center of mass. The coordinates (α,ŷ) of the center of mass of a lamina occupying the region D and having density function ρ(x,y) are: α= = ∬xρ(x,y)dA ŷ= = ∬yρ(x,y)dA My 1 __ 1 __ My __ __ m m m m

9. Moment of Inertia The moment of inertia of a particle of mass m about an axis is defined to be mr^2, where r is the distance from the particle to the axis. We extend this concept to a lamina with density function ρ(x,y) and occupying a region D by proceeding as we did for ordinary moments: we use the double integral: The moment of inertia of the lamina about the x-axis: Ix =y^2ρ(x,y)dA Similarly the moment about the y-axis is: Iy=x^2ρ(x,y)dA It is also of interest to consider the moment of inertia about the origin, also called the polar moment of inertia: I0=∬(x^2+y^2)ρ(x,y)dA Also notice the following: I0=Ix+Iy