Find the volume of the solid obtained by rotating the region

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SUMMARY

The volume V of the solid obtained by rotating the region bounded by the curves y = 5x and y = 5√x about the line y = 5 is calculated using the formula A(x) = π(R² - r²). The correct radii are R = 5 - 5x and r = 5 - 5√x, leading to the integral V = π∫₀¹(25x² - (10/3)x^(3/2))dx. The final volume is V = (21/3)π. The error in the initial attempt was not properly accounting for the axis of rotation and the correct setup of the radii.

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Gundown64
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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\sqrt{x} about y = 5

Homework Equations


A(x)=∏(R2-r2)

The Attempt at a Solution


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

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

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

V=∏(\frac{25}{3}-\frac{4}{3})
V= \frac{21}{3}∏

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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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\sqrt{x} about y = 5

Homework Equations


A(x)=∏(R2-r2)

The Attempt at a Solution


A(x)=∏(5x)2-(5\sqrt{x})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 - \frac{10}{3}x\frac{3}{2})

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

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

V=∏(\frac{25}{3}-\frac{4}{3})
V= \frac{21}{3}∏

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!
 
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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