What is the volume of a solid formed by rotating a region around the y-axis?

In summary, the conversation discusses finding the volume of a solid formed by revolving a region bound by the graphs of y = x^2 + 1, x = 0, and y = 10 around the y-axis. The homework equations used are pi r^2 and the attempt at a solution involves solving for y = 10 and using horizontal disks. The correct integral is found to be pi∫(y-1)dy from 1 to 10 by sketching graphs of the region and solid. The importance of sketching graphs is emphasized in order to avoid incorrect answers.
  • #1
Chas3down
60
0

Homework Statement


Find the volume of the solid formed by revolved the region bound by the graps of y =x^2 + 1, x=0, and y =10 around the y axis.


Homework Equations


pi r^2


The Attempt at a Solution



The x bound 0 is given to us.
I solved for y=10 by 10=x^2 + 1
I am not sure if I have to go from -3 to 0, and 0 to 3, or because the bound of 0 was give nto us, does that mean 0 is the lower bound? Also wasn't sure about the rotation around the y axis, I assumed the y-axis would be 0 so I didn't factor it into the equation..

3
pi ∫ (x^2 + 1)^2 dx = 348pi/5
0
 
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  • #2
Chas3down said:

Homework Statement


Find the volume of the solid formed by revolved the region bound by the graps of y =x^2 + 1, x=0, and y =10 around the y axis.


Homework Equations


pi r^2


The Attempt at a Solution



The x bound 0 is given to us.
I solved for y=10 by 10=x^2 + 1
I am not sure if I have to go from -3 to 0, and 0 to 3, or because the bound of 0 was give nto us, does that mean 0 is the lower bound? Also wasn't sure about the rotation around the y axis, I assumed the y-axis would be 0 so I didn't factor it into the equation..

3
pi ∫ (x^2 + 1)^2 dx = 348pi/5
0

This is incorrect. Did you draw a sketch of the region and one of the solid? It looks to me like you are thinking that the region is being revolved around the x-axis, not the y-axis.

This simplest way to do this, I believe, is to use horizontal disks.
 
  • #3
oh, thanks! I watched some videos and looked over my noted, and I think I have a better attempt at it now, I did everything in respects to y...

Limits from 1 to 10
Plugged in x=0 to y=x^2 + 1
y=10 was given to us

solved the equal y=x^2 +1 for x to get x = sqrt(y-1)End up with:

10
pi∫(y-1)dy
1
 
  • #4
Yes, that's the correct integral now. It's easy to evaluate.

It's important to sketch graphs, both of the region and the solid formed by revolving the region. Many students think they are saving time by skipping this step, but they wind up taking more time because of incorrect answers.
 
  • #5
Mark44 said:
Yes, that's the correct integral now. It's easy to evaluate.

Alright, thanks for not spoon feeding it, starting to get this volume rotation.
 
Last edited:

FAQ: What is the volume of a solid formed by rotating a region around the y-axis?

What is the volume of a rotated solid?

The volume of a rotated solid is the amount of space enclosed by the solid when it is rotated around a certain axis or line.

What is the formula for calculating the volume of a rotated solid?

The formula for calculating the volume of a rotated solid depends on the shape of the solid and the axis of rotation. For example, the formula for a cylinder rotated around its height axis is V = πr^2h, while the formula for a cone rotated around its slant height axis is V = (1/3)πr^2h.

How is the volume of a rotated solid different from the volume of a regular solid?

The volume of a rotated solid takes into account the additional space created by the rotation around an axis, while the volume of a regular solid only considers the space enclosed by its original shape.

What are some real-life applications of calculating the volume of a rotated solid?

The calculation of the volume of a rotated solid is used in engineering and architecture for designing and constructing structures such as bridges and buildings. It is also used in manufacturing processes for creating objects with specific volumes, such as bottles or pipes.

Are there any limitations to calculating the volume of a rotated solid?

The calculation of the volume of a rotated solid assumes that the solid is perfectly symmetrical and has a constant cross-section. This may not be the case in real-life scenarios, so the calculated volume may not be completely accurate.

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