This is about volume of solid of revolutions. I have reviewed my practice problems and completed them again. Can someone please check two problems for me. They are posted below in comments.

Question

anonymous · Answer

13. Volume = $$\int\limits_{0}^{6}[ \frac{ (-3)^{2}\pi \times6 }{ 3 }]dy$$
14. volume = $$\int\limits_{-4}^{6}[\pi(\frac{ -y-4 }{ 2 })^{2} - (3)^{2}]dx$$

anonymous · Answer

Please someone let me know if these are correct. I really want to make sure I understand this concept and I am doing these right.

anonymous · Answer

$$
\int_{-5}^{-2} 2 \pi  (6-(-2 x-4)) (-x-2) \, dx=18 \pi

$$

anonymous · Answer

This was for the first one

accessdenied · Answer

For (13), I observe that you used the volume of a cone formula with radius 3 and height 6 as the integrand.  However, this already will give you the volume you were looking for without integration

anonymous · Answer

the problem, you put above, is that 14 and if so how did you get that? Would you know how to do thirteen with integration?

anonymous · Answer

For the second one
$$
\int_{-5}^{-2} (6+3)^2 \pi  \, dx-\int_{-5}^{-2} \pi  ((-2 x-4)+3)^2 \,
   dx=243 \pi -117 \pi =126 \pi

$$

anonymous · Answer

@ganeshie8

anonymous · Answer

@ganeshie8 this is the ones that I have not gotten. The other ones I had gotten help. These are the last two that im hoping I got right but through the comments above I don't think they are right. I don't know how those aove got to those answers though

ganeshie8 · Answer

can u go back to ur old post where we had drawings ?

anonymous · Answer

no problem! just comment on which one you want to use and ill go to it(:

ganeshie8 · Answer

ive tagged u there

ganeshie8 · Answer

il tag u again