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?
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).
If you would graph the curve, and the rotation line, the problem would be obvious. If you rotate around the $y=7$ axis, the radius of rotation is along $x$ direction. So you need to calculate $\pi\int_0^2[x(y)]^2dy$, where $x(y)=y^2+1$. The lower limit is $y=0$, the upper limit is $(5-1)^{1/2}$.