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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.

$$y=\left(x-1\right)^{1/2}$$$$y=0$$$$x=5$$

rotate about the line y=7

I used this formula$$\int_1^5π\left[7-\left(x-1\right)^{1/2}\right]^{2}dx$$and got 388π/3, which is not correct. I changed the lower limit to 0 but it is also wrong. Can someone help me?

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  • $\begingroup$ You should reformulate the question. Be specific : which curve ? Rotation around which line ? Please edit the OP. $\endgroup$ – Adren Feb 13 '17 at 4:20
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I think your integral should be $\int_1^5 \pi 7^2 dx - \int_1^5 \pi \left(7 -(x-1)^{\frac{1}{2}}\right) ^2dx$.

If $A$ is the region bounded below by the horizontal line $y = 0$, above by the curve $f(x) = (x-1)^{\frac{1}{2}}$ and on the left and right by the vertical lines $x = 1$ and $x = 5$ respectively,...

and furthermore if $B$ is the region bounded above by the horizontal line $y = 7$, below by the curve $f(x)$, and on the left and right by the same vertical lines $x= 1$ and $x = 5$ as before,...

then $A$ and $B$ together form the rectangle $R$ bounded above and below by the horizontal lines $y = 7$ and $y = 0$ and bounded on the left and right by the vertical lines $x = 1$ and $x = 5$.

If you are asked to revolve $A$ about the axis of revolution $y = 7$. So you should revolve $R$ about this axis to form a cylinder (whose volume is the first integral in the difference written above), then remove from this cylinder the revolution of $B$ about the same axis (the volume of this revolution is the second integral above).

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