water partially fills a hemispherical bowl of radius 40cm so that the depth of the water is 20 cms. write down an integral representing the volume of the water in the bowl
To write down an integral representing the volume of the water in the bowl, we can use the concept of integration to calculate the volume of a solid when it is rotated around a specific axis.
In this case, we will integrate the cross-sectional areas of the water as it fills up the bowl. Since the bowl is a hemisphere, we can consider the cross-sectional area at a given height as a circle.
Let's denote the height of the water as 'h'. From the given information, we know that the depth of the water is 20 cm. Therefore, the height 'h' can be represented as (40 - 20) = 20 cm.
Next, we need to find the radius of the circle at height 'h'. Since the bowl is a hemisphere, the radius will change as we move up or down. We can use similar triangles to determine the radius at height 'h'.
Using the similar triangles:
radius at height 'h' / height 'h' = radius of the bowl / total height (40 cm)
Using this relationship, we can calculate the radius at height 'h' as follows:
radius at height 'h' = (radius of the bowl) * (height 'h' / total height)
Substituting the given values:
radius at height 'h' = 40 * (20 / 40) = 20 cm
Now that we have the radius at height 'h', we can write down the integral representing the volume of the water in the bowl.
Let's integrate the cross-sectional areas of the water from the bottom of the bowl (height 0) to the height 'h'. Since the bowl is symmetric, we can integrate only from 0 to 'h' and then multiply the result by 2.
The formula for the cross-sectional area of a circle is A = πr^2, where 'r' is the radius.
Therefore, the integral representing the volume of the water in the bowl is:
Volume = 2 * ∫[from 0 to h] (π * r^2) dh
Substituting the value of the radius at height 'h':
Volume = 2 * ∫[from 0 to 20] (π * (20)^2) dh
Simplifying,
Volume = 2 * π * 400 * ∫[from 0 to 20] dh
Now, integrating with respect to 'h',
Volume = 2 * π * 400 * [h] [from 0 to 20]
Evaluating the integral:
Volume = 2 * π * 400 * (20 - 0)
Simplifying,
Volume = 2 * π * 400 * 20
Volume = 16,000π cm^3
Therefore, the integral representing the volume of the water in the bowl is 16,000π cm^3.
To calculate the volume of the water in the hemispherical bowl, we need to integrate the cross-sectional area of the water with respect to the height.
Let's consider a small horizontal slice of the bowl at a certain height h from the bottom. The cross-sectional shape of this slice is a circular disc.
The radius of this disc can be calculated using the formula for a circle's radius at a given height from the center of the bowl:
r = √(radius^2 - height^2)
In this case, the radius of the bowl is 40 cm, so the radius of the circular disc at height h is:
r = √(40^2 - h^2)
The area of this circular disc can be calculated using the formula for the area of a circle:
A = π * r^2
Substituting the expression for r, we have:
A(h) = π * (√(40^2 - h^2))^2
Simplifying:
A(h) = π * (40^2 - h^2)
Finally, to find the volume of the water in the bowl, we integrate the cross-sectional area A(h) with respect to height h, from the bottom of the bowl (h = 0) to the depth of the water (h = 20):
V = ∫[0,20] π * (40^2 - h^2) dh
Thus, the integral representing the volume of the water in the bowl is:
V = ∫[0,20] π * (40^2 - h^2) dh