Find the volume of the solid obtained by rotating the region

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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!
 

Answers and Replies

  • #2
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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.
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
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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!
 

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