Integrals can be used any time we accumulate something. It could be accumulating speed to get distance, lengths to get area, even area to get volume or force to get work.

Consider a rectangular box, $$l\cdot w \cdot h$$. Notice that the $$l\cdot w$$ is the area of the base. So, Volume = Area of Base x height. Consider the integral:

$$! \int_0^h (l\cdot w) \, dx =l\cdot w \cdot h$$

This works for any area as long as the the top and bottom are parallel and the cross-sections have the same area.

$$! \int_0^h \textrm{Area}\,dx = Volume$$

#### Example 1

Consider the functions $$y=\sqrt{x}$$ and $$y=-\sqrt{x}$$. Imaging a square, perpendicular to the $$x-y$$-plane and $$x$$-axis whose diagonals span between these curves.

The length of any diagonal would then be $$2 \sqrt{x}$$ and a side length would be $$\sqrt{2x}$$ Now let that square slide from $$x=0$$ to $$x=4$$. Let’s sketch this solid and find it’s volume.

Volume = $$\displaystyle \int_0^4 \textrm{side}^2\,dx = \int_0^4 \big( \sqrt{2x}\big)^2\,dx = \int_0^4 2x\,dx = \big[x^2\big]_0^4 = 16$$ square units.

#### Consider two pipes intersecting

### Rotational Volumes

We can also find volumes by rotating areas or accumulating discs

Consider rotating the line $$y=3$$ about the $$x$$-axis.

To get a cross-sectional area we consider that a circular disc has an area of $$\pi r^2$$ and here the radius is the distance from the $$x$$-axis to the line $$y=3$$.

$$! \int_{x=0}^5 \pi (3)^2\,dx = \pi\cdot 3^2 \cdot 5$$

We can rotate any curve about an axis to generate volumes of all sorts.

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