a spherical fishtank has radius of 25 cm. The tank is filled with water with a depth of 20 cm. By using integral find the volume of the water inside the tank.
V=π∫ͪₒ(2Rx-x²)dx=π(Rx²- x³/3)|ͪ ₒ=π(Rh²- h³/3)= πh²(R-h/3)=....
To find the volume of the water inside the spherical fishtank, we can use the method of integration.
First, let's visualize the situation. We have a spherical fishtank with a radius of 25 cm. The water inside the tank has a depth of 20 cm. We want to find the volume of this water.
The volume of a solid of revolution can be calculated using the formula:
V = ∫[a,b] A(x) dx
where V is the volume, A(x) is the area of the cross-section at each value of x, and [a, b] is the interval over which we are integrating.
In our case, we can consider the cross-section of the water as a circle with variable radius r at each value of x. The radius r can be expressed in terms of x using the equation of a circle:
r = sqrt(R^2 - x^2)
where R is the radius of the fishtank and x represents the distance from the top of the water.
Since the depth of the water is 20 cm, the values of x will vary from -20 cm to the radius of the fishtank (25 cm). Thus, the interval for integration is [-20, 25].
Now, we need to calculate the area of the cross-section A(x) at each value of x. The area of a circle is given by:
A(x) = π * r^2
Substituting r = sqrt(R^2 - x^2) into the equation, we have:
A(x) = π * (R^2 - x^2)
where R is 25 cm.
Now, we can plug in the values into the volume formula:
V = ∫[-20,25] π * (R^2 - x^2) dx
V = π * ∫[-20,25] (R^2 - x^2) dx
V = π * ∫[-20,25] (625 - x^2) dx
V = π * [625x - (1/3)x^3] evaluated from -20 to 25
V = π * [(625(25) - (1/3)(25^3)) - (625(-20) - (1/3)(-20^3))]
V = π * [(15625 - 5208.33) - (-12500 + 5333.33)]
V = π * (10416.67 + 17833.33)
V = π * (28250)
V ≈ 88796.73 cm^3
Therefore, the volume of the water inside the spherical fishtank is approximately 88796.73 cm^3.