1 x4 and X 12. A parabolic curve y 16 axel area find X,), Ix, and...
296. Area under a curve. The area of the region bounded by the curve y = (-2<x< 2), the x-axis, V4 - x4 V4- and the lines x = a and x = b(a < b) is given by sin - €) - sin-"). a. Find the exact area if a 1 and 1 b. Find the exact area if a = -V3 and 5 = vā.
1) Problem 12 The area of the region bounded by the parabola x y-3) and the line y x is Problem 13 The base of a solid S is the parabolic region [(x.y):x s y S 1). Cross-sections perpendicular the y-axis are squares. Find the volume of the solid S
1) Problem 12 The area of the region bounded by the parabola x y-3) and the line y x is Problem 13 The base of a solid S is the...
Given the composite section below, calculate IX and IY 10" 5" 6" Calculations I.D. Area Ix+A d' Y A Y (n) A d d' in' dv (in) 1 (in' (in) 2. 3 Σ Y Ir Calculations I+ A.d (in A d'x dx A X Area X I.D. (in) (in' (in) (in) (in (in') (in) 1 3 Σ X=
Area between the curve problems
1. Find the area between y 1/x, y 1/x2 and x 2 2. Find the area between y 8-x2, y x2, x -3 and 3. 3. Find (approximately) the area between y r cos (2) and y
1. Find the area between y 1/x, y 1/x2 and x 2 2. Find the area between y 8-x2, y x2, x -3 and 3. 3. Find (approximately) the area between y r cos (2) and y
Find the surface area of the solid of revolution obtained by
rotating the curve
x=(1/12)(y^2+8)^(3/2)
from ?=2 to ?=5 about the x-axis:
(1 point) Find the surface area of the solid of revolution obtained by rotating the curve X= +8)3/2 from y = 2 to y = 5 about the x-axis:
Locate the centroid (x, y) of the shaded area. Then find Ix and Iy.Lifesaver given to correct answer with all work shown.
An area is defined by two curves y = x and y = x2 as shown below. (a) (2 pt) Define vertical and horizontal infinitesimal elements. (b) (1 pt) Find the total area. (c) (2 pts) Calculate the x- and y-coordinates of the centroid C. (d) (2 pts) Calculate area moments of inertia about x and y axes (Ix and Iy) first. (e) (2 pts) Apply the parallel axis theorem to find area moments of inertia about the centroidal axis...
(1 point) Find the length of the curve defined by
y=18(8x2−1ln(x))y=18(8x2−1ln(x))
from x=4x=4 to x=8
(1 point) Find the area of the region enclosed by the
curves:
2y=4x−−√,y=4,2y=4x,y=4, and 2y+1x=52y+1x=5
HINT: Sketch the region!
(1 point) Find the volume of the solid obtained by rotating the
region bounded by the given curves about the specified axis.
y=2+1/x4,y=2,x=4,x=9;y=2+1/x4,y=2,x=4,x=9;
about the x-axis.
(1 point) Find the length of the curve defined by y = $(8x? – 1 In(x)) from x = 4...
find the area between the curve y=x^2+1 and y=x(2-x). sketch a diagram of this area. for what value(s) of n does y=x^n satisfy the differential equation: x^2(d^2y/dx^2)+5x(dy/dx)-21y=0
show all work!!
Given: Area under parabolic curve as shown. Find: Centroid of area under curve. Im im d (1.0, 1.5) m (0.25, 0.10) m O (0.33, 0.20) m (0.75, 0.30) m