# 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 y**

^{2}**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 360

^{0}.
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 360

^{0}(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 x

^{2}, Y-axis, line y = 0, and y = 2 rotates around the Y axis of 360^{0}.
Answer:

Y = 1/2 x

^{2}ó x^{2}= 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 y

_{1}= f(x), y_{2}= 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 360_{0}, 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 = x

^{2}and y = x + 2 is rotated around the X-axis in one full rotation.
Answer:

First determine the
intersection of the two curves.

x

^{2}= x + 2
x

^{2}- 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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