Question

I need to find the volume of the solid obtained by rotating y=sqrt(x) and y=x^2 about the y axis using the washer method, need help with the steps

Answer 1

well; your bounds are 0 to 1 right?

Answer 2

\[2pi\int_{0}^{1}x(x^{1/2})-x(x^2)\ dx\]

Answer 3

does it not matter if it's rotated about the x axis or y axis?

Answer 4

well, this shape doesnt matter; but i posted the 'shell' method which tends to be simpler

Answer 5

they ask for the washer method, haha, this is what confuses me

Answer 6

the 'washer' method adds up areas of circles so it integrates pi r^2

the shell method adds up areas of sheets like paper; so the method works out easier, simpler

Answer 7

the washer rides up the y axis and its radius = x values tho i fyou wanna try that way

Answer 8

yeah I know, I know how to use the shell method, but they want me to answer using the washer or disk method in the first question, the second question asks to use the shell method with the very same functions

Answer 9

\[pi\int_{0}^{1}[x^2]^2-[x^{1/2}]^2 \ dx\]
\[pi\int_{0}^{1}[y^{1/2}]^2-[y^2]^2 \ dy\]

Answer 10

perhaps it's the same answer because the bounds are (0,0) and (1,1)

Answer 11

its the same answer no matter how hard they want you to make it lol

Answer 12

do they want just a number answer? or do you have to sit there ant type in the washer method formula?

Answer 13

k thanks a lot, so you just change the functions to x= and solve...they want me to draw the graph, draw the typical washer form and fill out the washer method formula

Answer 14

the radius for the washer are f(y) from 0 to 1, right?

Answer 15

they cross at 0 and 1 so that's right

Answer 16

http://www.wolframalpha.com/input/?i=int%28pi%28y+-+y^4%29%29dy+from+0+to+1

Answer 17

http://www.wolframalpha.com/input/?i=y%3Dx^2+AND+y%3Dsqrt%28x%29+from+0+to+1

Answer 18

Thanks again, big help