The double integral of over the rectangle is if this limit exists evaluate double integral

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY FACULTY OF APPLIED SCIENCE ******* REPORT CALCULUS CLASS: CC07 – GROUP INSTRUCTOR : PHAN THỊ KHÁNH VÂN SEMESTER 221 STUDENT PERFORMANCE Thái Huỳnh Anh Phương Huỳnh Trung Nghị STUDENT ID 2053353 2152799 Đặng Đức Trọng Trương Nhật Quang Nguyễn Đạt Minh Nguyễn Văn Khôi Nguyên Nguyễn Cường Thịnh 2115100 2153729 2053222 2151247 2153835 100% 80% 0% 0% 100% 100% 100% Work Assignment STT Name Student ID Part Details Huỳnh Trung Nghị 2152799 Double Intergral Theorem , Powerpoint Đặng Đức Trọng 2115100 Thái Huỳnh Anh Phương 2053353 Nguyễn Cường Thịnh 2153835 Trương Nhật Quang 2153729 Nguyễn Văn Khôi Nguyên 2151247 Nguyễn Đạt Minh 2053222 Aplication , Matlab , PDF Function several variables Theorem, Matlab Application , Matlab Derivative Theorem , Application TABLE OF CONTENTS Work Assignment .2 TABLE OF CONTENTS I DOUBLE INTERGRALS 1) Theorems 2) Application of double integral II DERIVATIVE 10 1) Theorems 10 2) Application of derivative 11 III FUNCTION SEVERAL VARIABLES 13 1) Theorems 13 2) Application of function several variable 20 THEOREM DEFINITION The double integral of over the rectangle is if this limit exists EVALUATE DOUBLE INTEGRAL We shall evaluate integral , where , then PROPORTIES: MORE THEOREM: APPLICATION OF DOUBLE INTERGRAL: A Caculate the area of region R Find the volume V of the solid S that is bounded by the elliptic paraboloid , the planes x=3 and y=3 , and the three coordinate planes Solution First notice the graph of the surface in Figure a and above the square region However , we need the volume of the solid bounded by the elliptic paraboloid , the planes x=3 and y=3 , and the three coordinate planes (a) The surface above the square region (b) The solid S lies under the surface above the square region Now let’s look at the graph of the surface in (b) We determine the volume V by evaluating the double integral over R2 : = clc; clear allall; fun = @(x,y) 27-2.*x.^2-y.^2 q = intergral2(fun,0,3,0,3); Fprintf('area of region = %gm2\n',q); B The mass and the center of mass Find the mass and the center of mass of the lamina that occupies the region D and has the given density function ρ D is the triangular region with vertices (0,0), (2,1) and (0,3); ρ(x, y) =x+y Solution clc; clear allall; fun1 = @(x,y) 1/6.*(x.*(x+y)) fun1 = @(x,y) 1/6.*(y.*(x+y)) ymin = @(x) x./2 ymax = @(x) 3-x q1 = intergral2(fun1,0,2,ymin,ymax); q2 = intergral2(fun1,0,2,ymin,ymax); fprintf("Centre of mass is = (%g, %g)",q1,q2); THEOREM PARTIAL DERIVATIVE SECOND ORDER PARTIAL DERIVATIVE TOTAL DERIVATIVES APPLICATION OF DERIVATIVE a The geometry meaning of partial derivative: Example: Given f(x,y) = - x2 – 2y2 Find f’x ,f’y at (1,1) and the geometry meaning of them Solution: f’x = -2x => f’’x (1,1) = -2 ; f’y = -4y => f’y (1,1) = -4 The graph of f is a paraboloid z = - x2 – 2y2 and the plane y=1 cut paraboloid by the parabol C1 = z = 2- x2, y=1 The slope of the tangent line T1 with parabol C1 at (1,1,1) is f’x (1,1) = -2 Similarly, the plane x =1 cut paraboloid by the parabol C2 : z = -2y2, x = The slope of the tangent line T2 with parabol C2 at (1,1,1) is f’y (1,1) = -4 clc; clear; syms x y z = - x^2 - 2*y^2; zx = diff(z, x); zy = diff(z, y); fx = subs(zx, {x,y}, {1,1}) fy = subs(zy, {x,y}, {1,1}) b The geometry meaning of derivative in a specific direction : Example: Find the direvative in the direction of vector which create an angle of with the line Ox at (1,2) of f(x,y) = x3 – 3xy + 4y2 Solution: f’x = 3x2 – 3y, f’y = -3x + 8y => f’(1,2) = f’x(1,2)cos + f’y(1,2)sin = ( 13-3 This is the speed of change of f(x,y) in direction clc; clear; syms x y z = x^3 - 3*x*y+ 4*y^2; zx = diff(z, x); zy = diff(z, y); fx = subs(zx, {x,y}, {1,2}) fy = subs(zy, {x,y}, {1,2}) f= cos(pi/6)*fx+sin(pi/6)*fy; disp(f) FUNCTION OF SEVERAL VARIABLES THEOREMS QUADRATIC SURFACES APPLICATION FUNCTION OF SEVERAL VARIABLES A.The range R of the trajectory 1.The range R of the trajectory of the projectile fired with initial velocity from the barrel made with the horizontal line at an angle is determined by the formula R = with g ~ 9.8 m/s Matlab code syms vo phi vo = input("Input the initial velocity"); phi = input("Input the angle"); g = 9.8 R = (v0^2 * sin(2*phi))/g B.Pythagorean problem Pythagorean theorem relating the hypotenuse c and the lengths of the legs a, b Solution: To find the hypotenuse by Pythagorean formula To find the length of leg by Pythagorean formula To find the length of leg by Pythagorean formula clc clear all syms choice disp("Pythagorean theorem for finding hypotenuse/leg") Choice = input("Enter you choice hypotenuse(0)/leg(1), input or 1: ") for choice = leg1 = input("input your first leg: ") leg2 = input("input your second leg: ") hypotenuse = sqrt(leg1^2+leg2^2) for choice = h = input("input your hypotenuse: ") leg = input("input your given leg: ") remainleg = sqrt(h^2-leg^2) C uniform accelerated motion problem Calculate displacement related to initial velocity, acceleration and time Solution: We have displacement formula in uniform acceleration motion With is the displacement is initial velocity is acceleration is time clc clear all syms v0 a t v0 = input("Input your inital velocity: ") a = input("Input your acceleration: ") t = input("Input your time: ") s = v0*t + (a*t^2)/2