Find the volume of the solid obtained by rotating the region

In summary, the volume V of the solid obtained by rotating the region bounded by y = 5x and y = 5√x about y = 5 can be found using the equation A(x)=∏(R2-r2), where R = 5 - 5x and r = 5 - 5√x. The relevant integral to solve is V=∏\int^{1}_{0}(25x2- \frac{10}{3}x\frac{3}{2})dx, which results in V= \frac{21}{3}∏. It is important to take into account the axis of rotation when determining the radii, as well as including the factor of
  • #1
Gundown64
9
0

Homework Statement


Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = 5x, y = 5[itex]\sqrt{x}[/itex] about y = 5

Homework Equations


A(x)=∏(R2-r2)

The Attempt at a Solution


A(x)=∏(5x)2-(5[itex]\sqrt{x}[/itex])2)
A(x)=∏(25 x2 - [itex]\frac{10}{3}[/itex]x[itex]\frac{3}{2}[/itex])

V=∏[itex]\int[/itex][itex]^{1}_{0}[/itex](25x2- [itex]\frac{10}{3}[/itex]x[itex]\frac{3}{2}[/itex])dx

V=∏([itex]\frac{25}{3}[/itex]x3-[itex]\frac{4}{3}[/itex]x[itex]\frac{5}{2}[/itex]) {0,1}

V=∏([itex]\frac{25}{3}[/itex]-[itex]\frac{4}{3}[/itex])
V= [itex]\frac{21}{3}[/itex]∏

Where did I go wrong? I can't figure out the about y=5. If I am correct, you would usually subtract 5 from the two radii, but since they intersect at y=5, the just flip on that intersection point and thus we don't need to find the lost area. Bad explanation, I know, maybe someone can explain it to me.

Thanks in advance!
 
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  • #2
Gundown64 said:

Homework Statement


Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = 5x, y = 5[itex]\sqrt{x}[/itex] about y = 5

Homework Equations


A(x)=∏(R2-r2)

The Attempt at a Solution


A(x)=∏(5x)2-(5[itex]\sqrt{x}[/itex])2)
You've made an error right off the bat. I hope you sketched the region being rotated.

You aren't taking into account the fact that the axis of rotation is the line y = 5. The larger radius, R, is 5 - 5x. The smaller radius, r, is 5 - 5√x.

Also, your relevant equation is missing a factor - Δx. If you substitute the values above for R and r, you should get the right result.
Gundown64 said:
A(x)=∏(25 x2 - [itex]\frac{10}{3}[/itex]x[itex]\frac{3}{2}[/itex])

V=∏[itex]\int[/itex][itex]^{1}_{0}[/itex](25x2- [itex]\frac{10}{3}[/itex]x[itex]\frac{3}{2}[/itex])dx

V=∏([itex]\frac{25}{3}[/itex]x3-[itex]\frac{4}{3}[/itex]x[itex]\frac{5}{2}[/itex]) {0,1}

V=∏([itex]\frac{25}{3}[/itex]-[itex]\frac{4}{3}[/itex])
V= [itex]\frac{21}{3}[/itex]∏

Where did I go wrong? I can't figure out the about y=5. If I am correct, you would usually subtract 5 from the two radii, but since they intersect at y=5, the just flip on that intersection point and thus we don't need to find the lost area. Bad explanation, I know, maybe someone can explain it to me.

Thanks in advance!
 
  • #3
Ok, that's what my question at the end was about. I wasn't sure if I was supposed to subtract the radii from 5. I'll try it out and see how it goes. Thanks!
 

1. What is the formula for finding the volume of a solid obtained by rotating a region?

The formula for finding the volume of a solid obtained by rotating a region is V = ∫abπ(R(x))^2dx, where a and b represent the limits of integration, R(x) is the radius of the solid at a given point, and π is the mathematical constant pi.

2. How is the region defined for finding the volume of a solid by rotation?

The region is defined by a continuous and closed curve in the x-y plane, with the x-axis as the axis of rotation. The curve must also be bounded by the limits of integration, a and b.

3. Can the region for finding the volume of a solid by rotation be rotated around the y-axis instead of the x-axis?

Yes, the region can be rotated around the y-axis by using the formula V = ∫abπ(R(y))^2dy, where R(y) is the radius of the solid at a given point along the y-axis.

4. What units are used for the volume of a solid obtained by rotating a region?

The units for volume will depend on the units used for the limits of integration and the radius function. Generally, the units will be in cubic units (such as cubic meters or cubic inches).

5. Are there any special cases to consider when using the formula for finding the volume of a solid by rotation?

Yes, there are special cases to consider, such as when the region being rotated is below the x-axis, or when the region is rotated around a different axis than the x or y-axis. In these cases, the formula for finding the volume may need to be adjusted accordingly.

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