Disk and Shell Method (Ignore my other thread)

[myanmar]

Reducing it to three questions, because I'm pretty confident on the others.
9. Find the volume of the solid generated by revolving about the line x = -1, the region bounded by the curves y = -x^2 + 4x - 3 and y = 0.
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I graphed everything, and then translated the graph 1 to the right making it y=-(x-1)^2+4(x-1)-3 and y=0 rotated around the y-axis because this is easier for me.

So then, I should be able to do this with the shell method, and get $$\int_2^4 2x \pi (-(x-1)^2+4(x-1)-3)\,dx$$

I don't think I'm right because this is negative
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10. Consider the region in the xy-plane between x = 0 and x=pi/2 bounded by y = 0 and y = sin x. Find the volume of the solid generated by revolving this region about the x-axis.
stuck here too

Honestly, I just am posting this because my answer seems to simple
$$\int_0^\pi/2 sin^2x\,dx$$

11. Let R be the region bounded by y = 1/x, y = x^2, x = 0, and y = 2. Suppose R is revolved around the x-axis. Set up but do not evaluate the integrals for the volume of rotation using: a) the method of cylindrical shells; b) the method of circular disks.
Honestly don't know where to start

 

[tiny-tim]

So then, I should be able to do this with the shell method, and get $$\int_2^4 2x \pi (-(x-1)^2+4(x-1)-3)\,dx$$

I don't think I'm right because this is negative
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10. …Honestly, I just am posting this because my answer seems to simple
$$\int_0^\pi/2 sin^2x\,dx$$

11. … Honestly don't know where to start

Hi myanmar!

9. No, it is positive … for example, if x = 3, the bracket is -4 + 8 - 3 = 1.

10. I assume you mean $$\int_0^{\pi/2} \pi sin^2x\,dx$$ ?

Yes, that's fine!

11. Don't know how to help you on this … you just have to draw the diagram carefully, and decide what the endpoints are of each slice.