Calculating Volume of Solid by Rotating Region About the x-Axis

  • Thread starter Painguy
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In summary: You have a small algebra mistake, though. You should get (8pi)/15.In summary, when rotating the region bounded by y = x^2 and the line y = 1 about the x-axis, the exact volume of the solid obtained is 4pi/5. However, there was some confusion with the given interval and the correct answer is actually 8pi/15.
  • #1
Painguy
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Homework Statement



Consider the region bounded by y = x^2 and the line y = 1. Find the exact volume of the solid obtained by rotating the region about the x-axis.

Homework Equations





The Attempt at a Solution


1=x^2
x=1
x=0

∫ π(1-x^2)^2dx from 0 to 1

(πx^5)/5 -(π2x^3)/3 +πx evaluated from 0 to 1

π/5 - π2/3 +π = (3π-10π+15π)=8π/15

This answer is wrong. What am I doing wrong?
 
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  • #2
Painguy said:

Homework Statement



Consider the region bounded by y = x^2 and the line y = 1. Find the exact volume of the solid obtained by rotating the region about the x-axis.

Homework Equations





The Attempt at a Solution


1=x^2
x=1
x=0

∫ π(1-x^2)^2dx from 0 to 1

(πx^5)/5 -(π2x^3)/3 +πx evaluated from 0 to 1

π/5 - π2/3 +π = (3π-10π+15π)=8π/15

This answer is wrong. What am I doing wrong?

x=0 isn't an x value where the two curves intersect. x=1 is. There's another one.
 
  • #3
Painguy said:

Homework Statement



Consider the region bounded by y = x^2 and the line y = 1. Find the exact volume of the solid obtained by rotating the region about the x-axis.

Homework Equations





The Attempt at a Solution


1=x^2
x=1
x=0

The lower limit is wrong. Graph the region.

∫ π(1-x^2)^2dx from 0 to 1

The integrand is wrong. Check your formula for the disk or washer method.

(πx^5)/5 -(π2x^3)/3 +πx evaluated from 0 to 1

π/5 - π2/3 +π = (3π-10π+15π)=8π/15

This answer is wrong. What am I doing wrong?
 
  • #4
i got my answer to be 4π/5

my integral was setup as 2pi ∫(x^2)^2 dx from -1 to 1

I'm not sure how I came about this answer.

If I use the washer method I end up with ∫ π(0)^2 - π(x^2)^2 from -1 to 1

∫ -π(x^4) dx from -1 to 1 the answer to this is the negative half of what I got. I did several other similar examples and they all displayed the same pattern. I'm still not sure what I'm doing wrong :/
 
  • #5
Painguy said:
i got my answer to be 4π/5

my integral was setup as 2pi ∫(x^2)^2 dx from -1 to 1

I'm not sure how I came about this answer.

If I use the washer method I end up with ∫ π(0)^2 - π(x^2)^2 from -1 to 1

∫ -π(x^4) dx from -1 to 1 the answer to this is the negative half of what I got. I did several other similar examples and they all displayed the same pattern. I'm still not sure what I'm doing wrong :/

Are you trying to figure out what's going on with examples in your notes? If so, why don't you look away from them and just try this one from scratch. It kind of looks like you are trying to find the volume of y=x^2 rotated around the x-axis. That would be the 'negative half' that you subtract from the volume of the cylinder you get from rotating y=1 around the x-axis. Is that it?
 
  • #6
Dick said:
Are you trying to figure out what's going on with examples in your notes? If so, why don't you look away from them and just try this one from scratch. It kind of looks like you are trying to find the volume of y=x^2 rotated around the x-axis. That would be the 'negative half' that you subtract from the volume of the cylinder you get from rotating y=1 around the x-axis. Is that it?
I'm not using any notes. The examples I'm getting are randomly generated online. I'm trying to get the volume of the solid in between y=1 and y=x^2 rotated around the x-axis. I'm guessing it should look something like this?:confused:

zfBf8.png


As for creating an actual expression I'm not sure how to approach this problem.

I guess the interval would be from -1 to 1 since y=1 and 1=x^2 x=+-1 and x=0 at x=0 y=0

I'll try again I suppose.
∫ pi(1^2)-pi(x^2)^2 dx

that ends up giving me 8pi/5. The actual answer is 4pi/5 which i can get if the interval is from 0 to 1. My guess is that the problem was improperly worded and forgot to mention x=0 as one of the bounds. Either way thanks for your help. I really appreciate it!
 
Last edited:
  • #7
That all seems ok to me.
 

1. What is the definition of volume?

The volume of a region is the amount of space that it occupies in three-dimensional space. It is typically measured in cubic units such as cubic meters or cubic centimeters.

2. How do you find the volume of a regular solid?

To find the volume of a regular solid, such as a cube or cylinder, you can use a formula that relates the dimensions of the shape. For example, the volume of a cube is calculated by multiplying the length, width, and height together (V = lwh).

3. What is the difference between volume and surface area?

Volume and surface area are both measurements of space, but they are not the same. Volume is the amount of space inside a solid figure, while surface area is the measurement of the outside surface of the figure.

4. How do you find the volume of an irregular shape?

To find the volume of an irregular shape, you can use the method of integration. This involves breaking down the shape into smaller, regular shapes and using calculus to find the volume of each shape. Then, you can add up the volumes to find the total volume of the irregular shape.

5. Can you find the volume of a liquid?

Yes, the volume of a liquid can be measured by using a graduated cylinder or other measuring container. The volume can be read off of the graduated markings on the container, which typically measure in milliliters or liters.

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