# Find the volume of the solid obtained by rotating the region

## 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

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.

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Mark44
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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

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.
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.