Calculus: (2lessons x 6- 12 problems). Lessons: Plane Areas (3prob) with trigonometric (2prob) polar coordinates (1prob) Solids of Revolution a Circular Disk Method (3problems) b. Cylindrical Shell Method (3problems) 2. List of books and reference materials where your solved problems came from 3. Do not submit examples from the reference books and solved problems from reviewers or the Internet as this will defeat the purpose of this assignment which is to encourage you to solve problems on your own and thus gain familiarity with the different Integration procedures.Plane Areas Problem 1: Area between two curves Problem: Find the area enclosed by y = candy = 4 - x2. Solution Plan: 1. Identify the points of intersection: 12 - 4- 22 2x2 - 4 - c'=2 - x-1V2 2. Set up the integral: A - [(4 - x?) - x3] dx - JV7(4 - 2x2 ) dx 3. Evaluate the integral: A = 4x - 2x3 V/2 3 A - 4V/2 - 2(V/2) -4V/2+ 2( -V2)3 3 A - 8\\/2 _ 8V/2 _ 24v/2-8v/2_ 16V/2 3 3 References: . Stewart, James. "Calculus: Early Transcendentals." Cengage Learning.Problem 2: Area between a line and a parabola Problem: Find the area enclosed by y - - andy = a + 2. Solution Plan: 1. Identify the points of intersection: x'- x+2 - x'-x-2-0 - (x-2)(2+1)-0 - :=2,-1 2. Set up the integral: A = [(x + 2) - x2] da 3. Evaluate the integral: 2 A = + 2x - A= 6+4 + 091 00 091 00 - 091 091- A = 2+4 + A = 6 - 091 00 - 3-8 + = + 2049 toles = - 29 6 References: . Thomas, George B., and Ross L. Finney. "Calculus and Analytic Geometry." Addison-Wesley.Problem 3: Area under a curve Problem: Find the area enclosed by y = " and the x-axis from a = -1 to x = 1. Solution Plan: 1. Set up the integral: A- Made 2. Evaluate the integral: A = References: . Larson, Ron, and Bruce Edwards. "Calculus." Cengage Learning.Trigonometric Functions Problem: Area between sine and cosine Problem: Find the area enclosed by y - sin(x ) and y = cos(x) from x - 0 to c = Solution Plan: 1. Set up the integral: A - J (cos(x) - sin(x)) da 2. Evaluate the integral: A = [sin(x) + cos(x)] A - sin () + cos () ] - [sin(0) + cos(0)] A = [1 + 0] - [0 + 1] =1-1=0 3. Interpret the result: The integral evaluates to 0 because the areas above and below the x-axis cancel each other out. To find the absolute area, consider the absolute values of the functions: A - Jo (cos(x) - sin(x)) da + 3 (sin(x) - cos(r)) da Evaluate the first integral: So (cos(x) - sin(x)) da = [sin(x) - cos(r)] - sin () - cos ()] - [0 - 1] - 42 - 12 +1=0+1-1 Evaluate the second integral: (sin(x) - cos(x)) dr = [- cos(x) - sin(r)] = [-1 -0] - -12 12 - 12 = -1 - [-v2] = -1+ v2\fThe area between the curves is 8.83 square units. Area Between y =sin(x) and y = cos(x) 1.0 0.8 0.6 - =sin(x) y=cos(x) 0.4 - 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 14 1.6