Find Volume of Rotated Region Bounded by x+3=4y-y^2 & x=0 About x-axis

In summary, the monument is 20 m high and has a horizontal cross-section that is an equilateral triangle. The area of the equilateral triangle is x^2. The volume of the monument is then
  • #1
Gauss177
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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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  • #2
Gauss177 said:

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
Gauss177 said:
*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
Gauss177 said:

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
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
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]
 

FAQ: Find Volume of Rotated Region Bounded by x+3=4y-y^2 & x=0 About x-axis

What is the formula for finding the volume of a rotated region?

The formula for finding the volume of a rotated region is ∫π(R(x))^2 dx, where R(x) represents the radius of the region at a given x-value.

How do you determine the boundaries of the integral for a rotated region?

The boundaries of the integral for a rotated region are determined by the points of intersection between the curves that define the region. In this case, the boundaries are found by solving the equations x+3=4y-y^2 and x=0 for their points of intersection.

What is the role of the x-axis in finding the volume of a rotated region?

The x-axis serves as the axis of rotation for the region. This means that the region will be rotated around the x-axis to create a three-dimensional solid, and the volume will be calculated based on this rotation.

Can this formula be used for any rotated region?

Yes, this formula can be used for any rotated region, as long as the boundaries of the region can be defined and the axis of rotation is known. It is a general formula that can be applied to various shapes and curves.

Is it necessary to use calculus to find the volume of a rotated region?

Yes, calculus is necessary to find the volume of a rotated region because it involves calculating the area of cross-sections of the solid at different intervals along the axis of rotation. This cannot be done using basic geometry alone.

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