Question

Find the volume of the solid generated by revolving the region in the first quadrant bounded by y=x^2+1 and y=0 about the x axis.

Answer 1

I keep getting 28pi/15 for the answer but apparently it is 8pi/15

Answer 2

I used the disc method \[\int\limits_{0}^{1}\pi(x^2+1)^2\]

Answer 3

\[\pi \int\limits_{0}^{1}x^4+2x^2+1\]

Answer 4

\[\pi(\frac{ x^5 }{ 5 }+\frac{2 x^3 }{ 3 }+x)\]

Answer 5

\[\pi \left[ \left( \frac{ 1 }{ 5 }+\frac{ 2 }{ 3 }+1 \right)-0 \right]\]

Answer 6

which was 28pi/15

Answer 7

Can anyone tell me where I'm going wrong?

Answer 8

who told you the bounds are from 0 to 1? there aren't in the question you posted

Answer 9

Am I using the wrong bound? That's the whole question.

Answer 10

is that the entire question though? because it doesn't state what the x bounds are, just the y bounds

Answer 11

That is the entire question

Answer 12

it can't be :-X lol

Answer 13

x^2 + 1 goes on forever

Answer 14

It is, haha. I'm studying past final exams and it's a multiple choice question

Answer 15

makes no sense at all. but you are doing it right. the problem must be the bounds

Answer 16

hmmm, this is frustrating. I did notice that for the final answer (3/15 + 10/15 + 15/15)=28/15....(3/15+15/15-10/15)=8/15....which is apparently the correct answer

Answer 17

I think I just may email my professor. It looks like there may be a typo. Maybe it's supposed to be -x^2+1? Then x could be + or -1

Answer 18

Thank you for looking at it for me

Answer 19

yes, 1 - x^2 makes a lot of sense. the bounds would be implied for this one and the answer is actually 8pi/15. great find

Answer 20

I bet that's the problem then. It looks like it got messed up in the printer or something because the equation is all on top of eachother so it's difficult to read.

Answer 21

ya :)