As a hardworking student, plagued by too much homework, you spend all night doing math homework. By 6am, you imagine yourself to be a region bounded by
y=5x2
x=0
x=3
y=0
As you grow more and more tired, the world begins to spin around you. However, according to Newton, there is no difference between the world spinning around you, and you spinning around the world. Unfortunately, you are so tired that you think the world is the x-axis. What is the volume of the solid you (the region) create by spinning about the x-axis?
I am not a fan of coating a simple question in cutesy wording, so here is the question the way it would appear in my assignment.
Find the volume when the region bounded by
y = 5x^2, x = 0 , x = 3 and y = 0 is rotated about the x-axis
V = π∫ 25x^4 dx from x = 0 to x = 3
= π [ 5x^5] from 0 to 3
= π(5(3^5) - 0)
= 1215π
Well, it seems like your tiredness has caused quite the mathematical mishap! But fear not, Clown Bot is here to put a smile on your face and shed some light on the situation.
To find the volume of the solid formed by spinning the region bounded by y = 5x^2, x = 0, x = 3, and y = 0 around the x-axis, we can use the method of cylindrical shells. This method involves calculating the volume of countless tiny cylindrical shells and then summing them up.
Now, let's get this comedy show on the road. Grab a pen and paper, and get ready to laugh... I mean, calculate!
First, we'll calculate the volume of one cylindrical shell. Imagine we're at a point (x, y) on the curve y = 5x^2. The radius of our cylindrical shell will be the x-value, and the height will be the y-value.
So, the volume of one cylindrical shell is given by:
V = 2πxy dx
Now, we have to express y in terms of x to get a neat little equation. The curve y = 5x^2 helps us with that.
V = 2πx(5x^2) dx
V = 10πx^3 dx
Now, integrating this equation from x = 0 to x = 3 will give us the total volume:
Total Volume = ∫(0 to 3) 10πx^3 dx
After integrating, we get:
Total Volume = πx^4 evaluated from 0 to 3
Total Volume = π(3)^4 - π(0)^4
Total Volume = 81π - 0
Total Volume = 81π units cubed
So, my mathematically-challenged friend, the volume of the solid you create by spinning around the x-axis is 81π units cubed. Now, go get some rest and dream of pi(e)!
To find the volume of the solid created by spinning the region bounded by the curves y=5x^2, x=0, x=3, and y=0 around the x-axis, we can use the method of cylindrical shells.
The volume of the solid can be calculated by integrating the circumference of each cylindrical shell multiplied by its height.
The height of each shell is given by the curve y=5x^2, and the circumference of each shell is given by 2πx.
The integral for the volume is:
V = ∫[a,b] (2πx * 5x^2) dx
where a is the lower limit and b is the upper limit.
In this case, a=0 and b=3.
V = ∫[0,3] (2πx * 5x^2) dx
To solve this integral, we can simplify the expression inside:
V = ∫[0,3] (10πx^3) dx
Now we can integrate:
V = 10π ∫[0,3] (x^3) dx
= 10π * [x^4/4] [0,3]
= 10π * [(3^4/4) - (0^4/4)]
= 10π * (81/4)
= 202.5π
Therefore, the volume of the solid created by spinning the region about the x-axis is 202.5π cubic units.
To find the volume of the solid formed by spinning the region bounded by the given equations around the x-axis, we can use the method of cylindrical shells.
The equation y = 5x^2 represents a parabola opening upwards. The region enclosed by y = 5x^2, x = 0, and x = 3 can be visualized as a shape resembling a cup. When this shape is revolved around the x-axis, it creates a solid with a hollow interior.
To determine the volume of this solid, we can integrate the area of infinitesimally thin cylindrical shells formed by revolving small strips of the region about the x-axis.
The radius of each cylindrical shell is given by the value of y (which is 5x^2), and the height of each shell is given by the infinitesimally small change in x, denoted as dx.
The volume element (dV) of each cylindrical shell can be calculated as: dV = 2πxydx, where x is the variable representing the position of the shell.
To find the total volume of the solid, we integrate this volume element over the range of x values from 0 to 3:
V = ∫[0,3] 2πxydx
Now, substituting y = 5x^2, we have:
V = ∫[0,3] 2πx(5x^2)dx
= 10π ∫[0,3] x^3 dx
Integrating x^3 with respect to x, we get:
V = 10π * [x^4/4] evaluated from 0 to 3
= 10π * [(3^4)/4 - (0^4)/4]
= 10π * (81/4)
= 810π/4
= 202.5π
Therefore, the volume of the solid created by spinning the given region about the x-axis is 202.5π cubic units.