WebWhat is the moment of inertia about the x-axis bounded by the curves y² = -16 (x-4), the line 8x-15y = 0 and the x-axis. A bee was flying upward along the curve that is the intersection … WebWhat is the moment of inertia about the x-axis bounded by the curves y² = -16 (x-4), the line 8x-15y = 0 and the x-axis. A bee was flying upward along the curve that is the intersection of z = x⁴+xy³+12 with the plane x=1. At the point (1,–2,5), itu event off on the tangent line.

calculus - Area bounded by the x-axis and the curve.

WebThe volume ( V) of a solid generated by revolving the region bounded by y = f (x) and the x ‐axis on the interval [ a, b] about the x ‐axis is. If the region bounded by x = f (y) and the y ‐axis on [ a, b] is revolved about the y ‐axis, then its volume ( V) is. Note that f (x) and f (y) represent the radii of the disks or the distance ... WebThe area between two curves and the area under a curve; The curve's average value; In Physics. Integrals are used to find: Centre of gravity; Mass and momentum of inertia of … cicak goreng

calculus - Finding the Area bounded by two lines and x-axis ...

WebThe semicircular region bounded by the curve x = 9 − y 2 and the y-axis is revolved about the line x = − 4. The integral that represents its volume is V = ∫ a b f (y) d y where a =, b =, and f (y) = See Example 4 page 284 for a similar problem. WebExample 2: If the ordinate x = a divides the area bounded by the curve y = (1 + [8 / x 2]), the x−axis and the ordinates x = 2, x = 4 into two equal parts, then what is the value of a? … WebJun 9, 2024 · #y=f(x)= sqrt x#, area of the region bounded by the graph of . #f(x)#, the x-axis and the vertical lines between #x=2 and x=5# is Area #A= int_a^b f(x)dx =int_2^5 x^(1/2) dx # or. #A= [ x^(1/2+1)/(1/2+1)]_2^5 = 2/3 [x^(3/2)]_2^5# or. #A=2/3 [5^(3/2)-2^(3/2)]=2/3(5 sqrt 5 - 2 sqrt 2)# or . #A~~ 5.57# sq.unit . Area under curve is #5.57(2 d p ... cicak anime