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integral
- Thread starter mathe25
- Start date
I assume you mean \(\displaystyle y=\sqrt{x+1}\). Be careful how you write it. Without LaTex you should have said y=sqrt(x+1).View attachment 2689 i have to find a volume(certain integral) of geometric body which is showed (produced) by y=sqrtx+1 rotation around y=-1...
That being said, what is the limiting value of x? The way the problem stands now the voulme would be infinite? What is the exact wording of the problem?
Looks like a prime time to use cylindrical shells.
HallsofIvy
Elite Member
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- Jan 27, 2012
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Personally, I have never liked "shells" or "washers" and you don't reall need them. A "generic" point on \(\displaystyle y= \sqrt{x+ 1}\), rotated around y= -1 will move in a circle with radius \(\displaystyle y-(-1)= \sqrt{x+1}- 1\). That circle with have area \(\displaystyle \pi r^2= \pi (\sqrt{x+1}- 1)^2\). Taking a "thickness" of dx, each disk has volume \(\displaystyle \pi(\sqrt{x+1}- 1)^2dx\). The total volume will be \(\displaystyle \pi\int_0^1(\sqrt{x}- 1)^2 dx\).
Now, that is the whole region from y= -1 up to \(\displaystyle y= \sqrt{x+ 1}\). To get the volume for the region from y= 0 up to \(\displaystyle y= \sqrt{x+1}\), you need to subtract the volume of the cylinder formed by the region from y=-1 to y= 0 (this is equivalent to the "washer" method). The volume of that cylinder is, of course, \(\displaystyle \pi r^2h= \pi\) since r= h= 1.
Now, that is the whole region from y= -1 up to \(\displaystyle y= \sqrt{x+ 1}\). To get the volume for the region from y= 0 up to \(\displaystyle y= \sqrt{x+1}\), you need to subtract the volume of the cylinder formed by the region from y=-1 to y= 0 (this is equivalent to the "washer" method). The volume of that cylinder is, of course, \(\displaystyle \pi r^2h= \pi\) since r= h= 1.
I have several differences from the previous answer, so I will repeat the same steps. For one thing, it looks to me like the shape is a solid, not with a hollow core.can u explain why is the total V=π int (sqrt x +1)^2 ann its not V=π int ((sqrt x+1))^2-1^2 ??? please ,i have it for tomorrow
The radius of a rotating point at (x,y) measured from the axis y=-1 is
\(\displaystyle r(x) = y(x) + 1 = \sqrt{x + 1} + 1 \)
and the area of the circle at x is
\(\displaystyle A(x) = \pi\ r^2 = \pi \ \left[\sqrt{x + 1} + 1\right]^2 = \pi \ \left[x + 2 + 2 \sqrt{x + 1}\ \right] \)
The volume element dV is A(x)dx. If the limits of x are [a,b], then
\(\displaystyle \displaystyle V = \int_a^b A(x)\ dx = \) · · · [I'm sure you can do the integration]
There is still a question about what the limits of integration should be. Only one is obvious from the statement of the problem: x=1 is a limit - but not clear if upper or lower.
If lower limit = 1, what is the upper?
If the upper limit is 1, perhaps the lower limit is x=-1, because the function does not exist for x<-1. The red curve on your plot extends to the point (-1,0) and no further. So I would use limits a=-1, b=+1.
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