Volume of solid

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Homework Statement


Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:

x+3=4y-y^2
x=0
About x-axis

Homework Equations



The Attempt at a Solution


I tried solving x+3=4y-y^2 for y, but I can't get it even though it seems simple.

*Edit*
Can someone check if I set up this integral correctly (for a different problem)?
Find volume of region bounded by given curve and rotated around given axis:
y=x^3
y=8
x=0
Rotate about x=2

Integral from 0 to 8 of: pi(2-0)^2 - pi(2-x^1/3)^2 dy

thanks
 
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Answers and Replies

  • #2
cristo
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Homework Statement


Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:

x+3=4y-y^2
x=0
About x-axis

Homework Equations



The Attempt at a Solution


I tried solving x+3=4y-y^2 for y, but I can't get it even though it seems simple.

y cannot be expressed as a function of x. Try plotting the graph, and you'll see that for each value of x there will be two values of y.

*Edit*
Can someone check if I set up this integral correctly (for a different problem)?
Find volume of region bounded by given curve and rotated around given axis:
y=x^3
y=8
x=0
Rotate about x=2

Integral from 0 to 8 of: pi(2-0)^2 - pi(2-x^1/3)^2 dy

thanks

Should this not be [tex] \pi\int_0^8 2^2-y^{2/3}dy [/tex]?
 
  • #3
Fredrik
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*Edit*
Can someone check if I set up this integral correctly (for a different problem)?
Find volume of region bounded by given curve and rotated around given axis:
y=x^3
y=8
x=0
Rotate about x=2

Integral from 0 to 8 of: pi(2-0)^2 - pi(2-x^1/3)^2 dy
I started with this one and right now I don't have time to do the other one. I think you can express this volume as

[tex]\int_0^8\pi (2-y^{1/3})^2 dy[/tex]

or equivalently as

[tex]\int_0^2 2\pi (2-x) x^3 dx[/tex]
 
  • #4
Fredrik
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Homework Statement


Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:

x+3=4y-y^2
x=0
About x-axis

Homework Equations



The Attempt at a Solution


I tried solving x+3=4y-y^2 for y, but I can't get it even though it seems simple.
First of all, you have to figure out what this region looks like. Define f by [itex]f(y)=-y^2+4y-3[/itex]. Calculate f´ and f". Solve the equation f´(y)=0 for y. Is f"(y) positive or negative when y is such that f´(y)=0? Find the points where the curve intersects the line x=0.

When you've done this you will know enough about what the region looks like to proceed. You have two options, either integrate over y (which means that you add up the volumes of cylindrical shells of thickness dy) or integrate over x (which means you add up the volumes of circular discs with a hole at the center). I think it's easier to integrate over y.

The solutions of the equation [itex]z^2+az+b[/itex] are [itex]z=-a/2\pm\sqrt{(a/2)^2-b}[/tex].
 
  • #5
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Thanks for help everyone. I have another question but didn't want to create another thread. I'm having trouble with these volume questions, so I hope some more examples will help me understand them better.

The height of a monument is 20 m. A horizontal cross-section at a distance x meters from the top is an equilateral triangle with side x/4 meters. Find the volume of the monument.
 
  • #6
Fredrik
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Do you know how to calculate the area of an equilateral triangle with side x/4? Call that A(x). The volume of a triangular slice of thickness dx is then A(x)dx. Integrate from 0 to 20.

[tex]\int_0^{20} A(x) dx[/tex]
 

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