Area and Volume (shell and disk method)

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Last few I can't seem to do, but on the others, can someone check my answers?. The last few I need help getting through.

1. Find the volume of the solid formed when the region bounded by the curves y = x^3 + 1, x = 1 and y = 0 is rotated about the x-axis.
My answer: int(pi(x^3+1)^2,x,-1,1)

2. Find the volume of the solid of revolution obtained by revolving the region bounded by y = 1/x and the lines x = pi/8 and x = pi/2 around the x-axis the x-axis.
My answer: int(pi(1/x)^2,x,pi/2,pi/8)

3. Find the area bounded by the curves f(x) = x^3 + x^2 and g(x) = 2x^2 + 2x.
int((x^3+1)-(2x^2+2x),x,-1,0) + int((2x^2+2x)-(x^3+1),x,0,2)

4. Find the area bounded by the curves y = sqrt(x), y = (5-x)/4, and y = (3x-8)/2
int(sqrt(x)-(5x-4)/4,x,1,3) + int(sqrt(x) - (3x-8)/2,x,3,4)

5. Let R be the region bounded by: y = x^3, the tangent to y = x^3 at (1,1), and the x-axis. Find the area of R integrating: a) with respect to x, and b) with respect to y.
Solving for tangent yields tangent = 3x-2
integrating for x yields
int(x^3-(3x-2),x,0,1)
b) Help me with this, not sure how to do at all.

6. Find the volume of the solid generated by revolving the region bounded by the graphs of y = x^2 - 4x + 5 and y = 5 - x about the line y = -1.
disk: int(pi((5-x)^2-(x^2-4x+5)^2),x,0,2)
shell: int(2pix(-x^2+3x),0,3)

7. Use cylindrical shells to find the volume of the solid obtained by revolving around the y-axis the region bounded by the curves y^2 = 8x and x = 2.
int(2pix(2sqrt(2x),x,0,2)

8. A cylindrical hole is drilled through the center of a sphere of radius R. Use the method of cylindrical shells to find the volume of the remaining solid, given that the solid is 6 cm high.
I believe the answer is 36 pi in^3, but can not seem to do this problem with the shell method.

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.

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.

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.
 
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Posting 11 questions at once (including 'check my work' problems) really isn't going to help you get a response. People who answer these want to focus on what your problem is, not the whole problem set. My eyes glazed over when I saw this. That said, the first one looks ok. Now I'll skip ahead to where you actually ask a question. Part 5a) isn't so good. The tangent line hits the x-axis at x=2/3. You can't do this one dx in a single integral, can you?
 
Since I can't seem to find the edit button, I'll post only what I need here. Ignore other problems

So, I'd appreciate some help on the following. Even one problem would be great.

In order of preference:
5, 10, 11, 9. (6&7 if possible)

5. (I figured out how to do b, but if #5a is wrong, I don't know where I went wrong.)

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.
a little stuck

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

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.
stuck.

6. & 7. I'd just like to know if my answers are right.
 
5a is wrong because the problem says 'bounded by the x-axis'. If the tangent line crosses the x-axis in the interval [0,1] then it's not the lower bound over the whole range of integration. Can we try and stick with one problem at a time?? Otherwise the confusion can be truly amazing.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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