Find the volume of the solid formed by revolving the region bounded by y=x^3, x=2 and y=1 about the y-axis.
Please help =(
To find the volume of the solid formed by revolving the region bounded by y=x^3, x=2, and y=1 about the y-axis, we can use the method of cylindrical shells.
Step 1: Sketch the given region
First, let's graph the region bounded by y=x^3, x=2, and y=1 in the xy-plane:
----------- y-axis
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|
| region
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x-axis
The region looks like a curve that starts at (1,1) and ends at (2,8).
Step 2: Determine the range of integration
To find the volume of the solid, we need to rotate this curve around the y-axis. The range of integration will be from y=1 to y=8, since those are the y-values that span the region.
Step 3: Setup the integral representing the volume
The volume of the solid can be calculated using the formula:
V = ∫ (2π * r * h) dy
In this case, the radius (r) is the x-value of the curve, which is x = (y)^(1/3), and the height (h) is the width of the cylindrical shell, which is dy.
So, the volume integral becomes:
V = ∫ (2π * (y)^(1/3) * dy) from 1 to 8.
Step 4: Evaluate the integral
Now we need to solve the integral:
V = 2π * ∫ (y^(1/3)) dy from 1 to 8
To integrate y^(1/3), we add 1 to the exponent and divide by the new exponent:
∫ (y^(1/3)) dy = (3/4) * y^(4/3)
Substituting the limits of integration:
V = 2π * [(3/4) * (8^(4/3)) - (3/4) * (1^(4/3))]
Simplifying:
V = 2π * [(3/4) * 32 - (3/4) * 1]
V = 2π * [(3/4) * 32 - (3/4)]
V = 2π * [(3/4) * 31]
V = 2π * (93/4)
V = 93π/2
So, the volume of the solid formed by revolving the region bounded by y=x^3, x=2, and y=1 about the y-axis is (93π/2) cubic units.
Did you make a sketch?
I would calculate the volume of a cylinder with radius 2 and height of 1, then subtract the region bounded by the y-axis, y= x^3 and y = 1
that volume is
[integral] (πx^2)dy from 0 to 1
= π[integral] y^(2/3) dy from 0 to 1
= π ((3/5)y^(5/3) | from 0 to 1
= π(3/5 - 0) = 3π/5
So volume = π(2)^2(1) - 3π/5 = 4π - 3π/5 = 17π/5
check my arithmetic