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Can anyone help? I've tried disc and shell method but the disc seems the most likely. What I can't seem to do is specify everything correctly.

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Hi Ian. In its current form, your question might not receive many answers. Please take a look at the How to Ask-page and try to improve your question according to the guidance found there. This may require you to show some effort on your part in terms of attempting a solution. –  Elias Dec 17 '12 at 16:33
See curvebank.calstatela.edu/volrev/volrev.htm for help. –  Elias Dec 17 '12 at 16:34

1 Answer 1

Maybe this will help:

  • Sketch the line $y=6$ and the region, $R$. Note that $R$ is a "half disc" shape in the first quadrant whose "diameter" (flat side) coincides with the interval $[0,4]$ on the $x$-axis.
  • Imagine what the solid looks like. You are revolving $R$ about the line $y=6$. This will generate a donut shape centered about the line $y=6$.
  • Indeed the disc method is appropriate. Now select an $x\in [0,4]$ and draw the vertical line segment in $R$ corresponding to $x$. Call this line segment $L_x$.
  • What shape do you get when $L_x$ is revolved about the line $y=6$?
    Answer: a disc, $D_x$, of outer radius $6$ and inner radius $6-(4x-x^2)$. So the area of $D_x$ is $$\text{area}(D_x)=\pi\bigl(6^2- (6-(4x-x^2) )^2 \bigr).$$
  • Finally, use the formula: $$\text{Volume}=\int_0^4 \text{area}(D_x)\,dx =\int_0^4 \pi\bigl( 6^2-(6-(4x-x^2) )^2 \bigr)\,dx.$$
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Thank you David, the explanation is excellent. I understand why the inner radius is 6, but my reckoning says the outer radius would be 6+(4x-x^2). What is wrong with my understanding? –  Ian Dec 17 '12 at 17:48
@Ian Sorry; I was off. The disc is in the first quadrant. The outer radius is $6$ and the inner radius is $6-(4x-x^2)$. I edited my answer to reflect this. –  David Mitra Dec 17 '12 at 18:23
I get it now. Your help has moved on my understanding of finding volumes with integrals a few more steps. Thank you. –  Ian Dec 17 '12 at 18:44
@Ian You're welcome. –  David Mitra Dec 17 '12 at 18:45

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