Ok the question is as follows:
Use cylindrical shells to find the volume of the solid that is generated hen the region that is enclosed by y=x^3, y=1, and x=0 is revolved about the line y=1.
For my integral I have 2pi * integral from 0 to 1 of [(y^(1/3))*(1-y)]
after a bunch of calculations I got a final answer of 9pi/14.
Is this correct??? If not can you please tell me the answer with work? Thank-you
Use cylindrical shells to find the volume of the solid that is generated hen the region that is enclosed by y=x^3, y=1, and x=0 is revolved about the line y=1.
For my integral I have 2pi * integral from 0 to 1 of [(y^(1/3))*(1-y)]
after a bunch of calculations I got a final answer of 9pi/14.
Is this correct??? If not can you please tell me the answer with work? Thank-you
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Yeah thats exactly what I got as well. Id say its correct. Im homeschooled in Calculus 1 and my book just taught me the shell method a couple weeks ago. I like it more than the disc method.
These problems can be difficult to set up. And once they are set up correctly, you need to pay extreme attention to detail.
I didnt see you wanted work. Here it is.
Okay so you are integrating with respect to y, so we have y from 0 to 1 on the integral sign. We need the radius of each shell, r, and we need the height of each shell, which is going to be x. You can easily see this by looking at the graph of y=x^3.
Thus our integral will look like this:
∫0_1(2*pi*x*r) dy
Since its dy we need everything in terms of y.
Since y=x^3, x=y^(1/3).
The integral becomes:
(moving the constant 2*pi outside the integral symbol)
2*pi*∫0_1(y^(1/3)*r) dy
Now we need the radius of each shell. Since each shell is revolved around y=1, the radius of each shell is going to be r=1-y.
This can easily be shown to be correct, considering that at y=1 the radius is 0 because were revolving the solid around y=1. And at y=0 radius is 1. That makes sense with our equation for r, since 1-0 is 1.
Okay so our final integral is this:
2*pi*∫0_1(y^(1/3)*(1-y)) dy
Using the distributive property:
2*pi*∫0_1(y^(1/3)-y^(4/3)) dy
Now we use the power rule:
2*pi*[ (3/4)*y^(4/3)-(3/7)*y^(7/3) ]0_1
2*pi*{(3/4)(1)-(3/7)(1)} - 2*pi*{0}
(when 0 is used, both terms will equal 0}
Next we subtract and simplify.
2*pi*{(21/28 - 12/28)}
2*pi {9/28}
18*pi/28
9*pi/14
You are correct.
These problems can be difficult to set up. And once they are set up correctly, you need to pay extreme attention to detail.
I didnt see you wanted work. Here it is.
Okay so you are integrating with respect to y, so we have y from 0 to 1 on the integral sign. We need the radius of each shell, r, and we need the height of each shell, which is going to be x. You can easily see this by looking at the graph of y=x^3.
Thus our integral will look like this:
∫0_1(2*pi*x*r) dy
Since its dy we need everything in terms of y.
Since y=x^3, x=y^(1/3).
The integral becomes:
(moving the constant 2*pi outside the integral symbol)
2*pi*∫0_1(y^(1/3)*r) dy
Now we need the radius of each shell. Since each shell is revolved around y=1, the radius of each shell is going to be r=1-y.
This can easily be shown to be correct, considering that at y=1 the radius is 0 because were revolving the solid around y=1. And at y=0 radius is 1. That makes sense with our equation for r, since 1-0 is 1.
Okay so our final integral is this:
2*pi*∫0_1(y^(1/3)*(1-y)) dy
Using the distributive property:
2*pi*∫0_1(y^(1/3)-y^(4/3)) dy
Now we use the power rule:
2*pi*[ (3/4)*y^(4/3)-(3/7)*y^(7/3) ]0_1
2*pi*{(3/4)(1)-(3/7)(1)} - 2*pi*{0}
(when 0 is used, both terms will equal 0}
Next we subtract and simplify.
2*pi*{(21/28 - 12/28)}
2*pi {9/28}
18*pi/28
9*pi/14
You are correct.