Definite integral! The volume of the body of revolution, sine function, within 0 to 2 pies, rotates one circle around a straight line parallel to the Y axis (such as x = - Pie) Definite integral! The volume of a body of revolution, sine function, is the volume of a body of revolution within 0 to 2 pies, rotating about a straight line parallel to the Y axis (e.g. x = - Pie)

Definite integral! The volume of the body of revolution, sine function, within 0 to 2 pies, rotates one circle around a straight line parallel to the Y axis (such as x = - Pie) Definite integral! The volume of a body of revolution, sine function, is the volume of a body of revolution within 0 to 2 pies, rotating about a straight line parallel to the Y axis (e.g. x = - Pie)

The simple method is to find out the area of an arch by using the second theorem of gurukin, and then imitate the volume formula of a ring, that is, the area of the cross-section circle multiplied by 2 π L is equivalent to straightening the big ring into a cylinder, and its height is 2 π L. L is the distance from the center of the cross-section circle to the center of the ring, because it is around the Y axis, the center of the pendulum line must be on the central axis, The volume of a body of revolution generated by the rotation of a figure area a around a fixed line L which does not intersect with it is equal to the product of the circumference of the circle through which the area a and its center of gravity pass. DX = a (1-cost) DT, s = ∫ [x1, X2] f (x) DX = ∫ [0,2 π] a (1-cost) * a (1-cost) DT = a ^ 2 ∫ [0,2 π] a (1-cost) *, In the case of [0,2 π] --2a-2a ^ 2sint [0,2 π] + (a ^ 2 / 2 / 2) [0,2 π [0,2 π] + (a ^ 2 / 2 / 2) [0,2 π] (1 + cos2t) DT = 2 π a ^ 2 (1 + cos2t) DT = 2 π a ^ 2 (1-2cost) + (a ^ 2 / 2 / 2 / 2 / 2) (2 π - 2) (2 π - 0) + (a ^ 2 / 4) sin 2T [0,2 π [0,2 π [0,2 π] in this paper, when t = π is the symmetry axis of the cycloid, the center is on it, the distance from Y axis to y axis is π a, the distance from the center to y axis is π a, the distance to y axis is π a, the distance to y axis is π a, the distance from Y axis is π a, the big* (3 π a ^ 2) = 6 π ^ 3A ^ 3

How to find the volume of the body of revolution (with definite integral) generated by the rotation of the figure enclosed by y = x & # 178; and x = y & # 178?

V = π∫ (0 to 1) [the fourth power of Y-Y]

It is known that the image of the first-order function can be obtained by the translation of the line y = 2x, and the area of the triangle enclosed by the line y = - 2x and the x-axis is 4
The intercept on the Y-axis and the area of the triangle formed by its image and the coordinate axis

On the y-axis, the intercept is 4 √ 2, and the area of the triangle is 8

Given that the image of the first-order function y = KX + B is parallel to y = - 2x + 1 and passes through the point (2, - 1), then the area of the triangle bounded by the image of the first-order function and the coordinate axis is

The image of the linear function y = KX + B is parallel to y = - 2x + 1
k=－2
X = 2, y = - 1
-1=-2×2＋b
b=3
y=－2x＋3
The intersection axis is at (0,3), (1.5,0)
The area of the triangle is # 189; × 3 × 1.5 = 2.25