Volume by Rotation Using Integration

In our daily lives, many of us encounter rotating objects such as flower vases, lamp shades, buckets, cans and so on. A question disturbs us, can we calculate the volume of the object. It's not easy, but with the Integral method we will learn below, it can answer that question.

Look at the buildings below and the swivel object produced when the buildings are rotated around the X-axis by 3600 (Figure-1).


The shape of the rotating object that occurs is as follows,


Volume by rotating objects surrounding the X-axis.


Note the area which is bounded by the curve y = f (x), the line x = a and line x = b which rotates around the X axis of 3600 (see Figure-2).


To get the volume of the rotating object, small tube pieces are made as an approach to the volume of the object as shown in Figure-3 (a). Note one of the tube pieces released in Figure-3 (b).


If the radius of the tube is y and the height is δx, then

δV = π y2 δx

The volume of the rotating object is the number of pieces of the tubes, so that


with n is the number of pieces of tube.
For δx which is quite small, a perfect volume approach will be produced, i.e.


The form of the limit above can be stated using integration as follows,


Example 1.
Calculate the volume of an object if the area is limited by the curve y = 3x + 2, the X-axis, the line x = 1 and x = 4 rotates around the X-axis of 3600.
Answer:




Volume by rotating objects surrounding the Y-axis.

Not only can we determine the volume of a rotating object around the X-axis, but it can also determine the volume of a rotating object in a plane if it rotates around the Y-axis. Therefore, observe the area which is limited to the x = f(y) curve, y = a, and y = b rotates around the Y-axis by 3600 (see Figure 5).


In the same way as when determining the volume of a rotating object in a plane when rotating around the X-axis, we get the formula for δV, namely:
With n is the number of tube pieces.
For δy which is quite small, a perfect volume approach will be produced, namely:
The form of the limit above, can be expressed as an integral form as follows.
Example 2.
Calculate the volume of the rotating object that occurs if the area is bounded by the curve y = 1/2 x2, Y-axis, line y = 0, and y = 2 rotates around the Y axis of 3600.
Answer:
Y = 1/2 x2 ó x2 = 2y


So, the volume is 4π volume units.

Finding the volume of the area between two curves when rotated about the X-axis

For example, f and g are continuous and nonnegative functions such that f(x) g(x) for [a, b]. And L is the area bounded by y1 = f(x), y2 = g(x), line x = a, and line x = b as in Figure 7. If the area is rotated around the X-axis as far as 3600, then the volume of the object occurs 
can be stated in the following form,


Example 3.
Calculate the volume of the rotating object that occurs if the area bounded by the curve y = x2 and y = x + 2 is rotated around the X-axis in one full rotation.

Answer:
First determine the intersection of the two curves.
x2 = x + 2
x2 - x – 2 = 0
(x – 2)(x + 1) = 0
x = 2 atau x = -1

so, the boundaries of the area are x = 2 and x = -1, so the intended volume is


So, the volume is 14.4 π volume units.





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