Show that the length of the curve y = sqrt((1 - x2/3))3 for 0 x 1 is 3/2.

Welcome back RUDDY — missed you this week. Here are the solutions to the questions. Question 1 Given the curve y = sqrt((1 - x^2/3))^3 for 0 x 1. i) Show that the length of the curve C is 3/2. Step 1: Rewrite the curve equation. The given curve is y = (1 - x^2/3)^3/2. To find the arc length, we need to calculate (dy)/(dx). Step 2: Differentiate y with respect to x. (dy)/(dx) = (3)/(2) (1 - x^2/3)^1/2 · (-(2)/(3) x^2/3 - 1) (dy)/(dx) = (3)/(2) (1 - x^2/3)^1/2 · (-(2)/(3) x^-1/3) (dy)/(dx) = -x^-1/3 (1 - x^2/3)^1/2 Step 3: Calculate ((dy)/(dx))^2. ((dy)/(dx))^2 = (-x^-1/3 (1 - x^2/3)^1/2)^2 ((dy)/(dx))^2 = x^-2/3 (1 - x^2/3) ((dy)/(dx))^2 = x^-2/3 - 1 Step 4: Set up the arc length integral. The arc length L is given by the formula: L = _a^b sqrt(1 + ((dy)/(dx))^2)\,dx Substitute ((dy)/(dx))^2: L = _0^1 sqrt(1 + (x^-2/3) - 1)\,dx L = _0^1 sqrt(x^-2/3)\,dx L = _0^1 x^-1/3\,dx Step 5: Evaluate the integral. L = [x^-1/3 + 1-1/3 + 1]_0^1 L = [x^2/32/3]_0^1 L = [(3)/(2) x^2/3]_0^1 L = (3)/(2) (1^2/3 - 0^2/3) L = (3)/(2) (1 - 0) L = (3)/(2) The length of the curve C is indeed 3/2. ii) The point (sqrt(3), sqrt((1-t^3)^3)) lies on the curve C. This part of the question seems to be incomplete or incorrectly transcribed. The given curve is y = sqrt((1 - x^2/3))^3. If a point (sqrt(3), sqrt((1-t^3)^3)) lies on the curve, then x = sqrt(3) and y = sqrt((1-t^3)^3). Substituting x = sqrt(3) into the curve equation: y = sqrt((1 - (3))^2/3)^3 This does not directly relate to t. Without further context or a complete question, it is not possible to provide a solution for this part. Assuming it asks to find t if the point lies on the curve, we would need to equate the y-coordinates. However, the x-value sqrt(3) is outside the domain 0 x 1. Therefore, this point cannot lie on the curve C as defined. iii) The curve C is rotated completely about the x-axis. Find the area of the surface of revolution obtained. Step 1: Recall the formula for the surface area of revolution about the x-axis. S = _a^b 2 y sqrt(1 + ((dy)/(dx))^2)\,dx Step 2: Substitute the known values. We have y = (1 - x^2/3)^3/2 and sqrt(1 + ((dy)/(dx))^2) = x^-1/3. S = _0^1 2 (1 - x^2/3)^3/2 x^-1/3\,dx Step 3: Use a substitution to evaluate the integral. Let u = 1 - x^2/3. Then du = -(2)/(3) x^-1/3\,dx. So, x^-1/3\,dx = -(3)/(2)\,du. Change the limits of integration: When x = 0, u = 1 - 0^2/3 = 1. When x = 1, u = 1 - 1^2/3 = 0. Substitute into the integral: S = _1^0 2 u^3/2 (-(3)/(2))\,du S = -3 _1^0 u^3/2\,du To reverse the limits, change the sign: S = 3 _0^1 u^3/2\,du Step 4: Evaluate the integral. S = 3 [u^3/2 + 13/2 + 1]_0^1 S = 3 [u^5/25/2]_0^1 S = 3 [(2)/(5) u^5/2]_0^1 S = 3 ((2)/(5) (1)^5/2 - (2)/(5) (0)^5/2) S = 3 ((2)/(5) - 0) S = (6)/(5) The area of the surface of revolution obtained is (6)/(5). iv) Using a theorem of Pappus, find the y-coordinates of a centroid of the arc C. Step 1: Recall Pappus's Second Theorem. Pappus's Second Theorem states that the surface area S of a surface of revolution generated by rotating a plane curve C about an external axis is equal to the product of the arc length L of C and the distance traveled by the centroid of C. S = 2 y L where y is the y-coordinate of the centroid of the arc C, and 2 y is the distance traveled by the centroid. Step 2: Use the values calculated in previous parts. From part (i), the arc length L = (3)/(2). From part (iii), the surface area S = (6)/(5). Step 3: Solve for y. (6)/(5) = 2 y ((3)/(2)) (6)/(5) = 3 y Divide both sides by 3: y = (6)/(5 · 3) y = (2)/(5) The y-coordinate of the centroid of the arc C is (2)/(5). Question 2 A square of side 2a is rotated through 2 about a line in its plane which is distant 3a from the centre O of the square. Use theorems of Pappus to find: a) The Volume of the Solid swept out by the Square. Step 1: Identify the properties of the square. Side length of the square is s = 2a. Area of the square A = s^2 = (2a)^2 = 4a^2. The centroid of the square is its geometric center. Step 2: Determine the distance of the centroid from the axis of rotation. The distance from the center O of the square to the axis of rotation is given as d = 3a. According to Pappus's First Theorem, the volume V of a solid of revolution generated by rotating a plane area A about an external axis is given by: V = 2 d A where d is the distance of the centroid of the area from the axis of rotation. Step 3: Calculate the volume. V = 2 (3a) (4a^2) V = 24 a^3 The volume of the solid swept out by the square is 24 a^3. b) The area of the Surface of the Solid. Step 1: Identify the perimeter of the square. The perimeter of the square P = 4s = 4(2a) = 8a. The centroid of the perimeter of the square is also its geometric center. Step 2: Determine the distance of the centroid from the axis of rotation. The distance from the centroid of the perimeter to the axis of rotation is d = 3a. According to Pappus's Second Theorem, the surface area S of a surface of revolution generated by rotating a plane curve (in this case, the perimeter of the square) about an external axis is equal to the product of the length of the curve P and the distance traveled by the centroid of the curve. S = 2 d P Step 3: Calculate the surface area. S = 2 (3a) (8a) S = 48 a^2 The area of the surface of the solid is 48 a^2. Send me the next one 📸