Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – MathOverview Dec 17 '12 at 16:33
See for help. – MathOverview Dec 17 '12 at 16:34

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: $$\text{Volume}=\int_0^4 \text{area}(D_x)\,dx =\int_0^4 \pi\bigl( 6^2-(6-(4x-x^2) )^2 \bigr)\,dx.$$
share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.