5.2.1 vol of sin x^2 ; 0\le x \le \dfrac{\pi}{2}

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In summary, the given conversation discusses the integral $\displaystyle \int_0^{\pi/2} \sin(x^2) dx$ and how it represents area, not volume. The speaker mentions that the statement of "volume" is incomplete and asks for clarification on what type of volume is being referred to. They also mention that this integral cannot be integrated in terms of elementary functions but can be approximated to any desired accuracy.
  • #1
karush
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volume of the solid $y=\sin (x^2)\quad 0\le x \le \dfrac{\pi}{2}$
$\displaystyle \int_0^{\pi/2}\sin (x^2)\ dx$
ok think this should be area not volume but hope my int is set up ok
 
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  • #2
The integral you have is indeed area. Your statement of “volume” is incomplete.
Is this a volume of rotation with respect to a given axis of rotation, or is it a volume
of similar cross sections whose base lies in a region defined by the given curve,
or … something else?
 
  • #3
If you really intend $\int_0^{\pi/2} \sin(x^2) dx$, that cannot be integrated in terms of elementary function but can be approximated to any desired accuracy.
 

1. What is the meaning of "5.2.1 vol of sin x^2 ; 0\le x \le \dfrac{\pi}{2}"?

The expression "5.2.1 vol of sin x^2 ; 0\le x \le \dfrac{\pi}{2}" represents the volume of a three-dimensional shape formed by rotating the curve y = sin x^2 around the x-axis, with x values ranging from 0 to pi/2.

2. How is the volume of this shape calculated?

The volume of this shape can be calculated using the formula V = π∫(f(x))^2 dx, where f(x) is the function being rotated and the integral is taken over the given interval.

3. What is the significance of the interval 0\le x \le \dfrac{\pi}{2} in the expression?

The interval 0\le x \le \dfrac{\pi}{2} indicates the range of x values for which the function is being rotated. In this case, it means that the curve y = sin x^2 is being rotated from x = 0 to x = pi/2, or in other words, a quarter of a full rotation.

4. Can this expression be used to find the volume of any shape?

No, this expression is specific to the given function and interval. The formula for calculating the volume of a three-dimensional shape formed by rotating a curve around an axis depends on the function and the interval of rotation.

5. What applications does this expression have in real-world scenarios?

This expression can be used to find the volume of objects with curved surfaces, such as vases or bottles. It can also be applied in physics and engineering to calculate the volume of rotating objects, such as gears or turbines.

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