Question

please help me!
http://answers.yahoo.com/question/index;_ylt=AgOQpn2oHp1rjlOOJ_X9Y2rsy6IX;_ylv=3?qid=20120320195205AA6t5wt
find the volume of the solid that results when the region bounded by x=(√5)y2 and the y-axis from y=-1 to y=1 is revolved around the y-axis

Answer 1

\[x=\sqrt{5}y ^{2}\]
correct

Answer 2

What is the volume?
I don't understand my homework. Number 2, 4, 5, 10, 11, 12 form practice problem set 26
please, help me!!

http://books.google.com.co/books?id=5-SisV1sLm0C&pg=PA218&lpg=PA218&dq=find+the+volume+of+the+solid+that+results+when+the+region+bounded+by+x%3D( √5)y2+and+the+y-axis+from+y%3D-1+to+y%3D1+is+revolved+around+the+y-axis&source=bl&ots=9nYt8kYaIk&sig=9ap_aYFgZw96zBO4LEZihUwABRM&hl=es-419&sa=X&ei=Qj9pT73TH6mlsALKkL2LCQ&ved=0CB0Q6AEwAA#v=onepage&q=find%20the%20volume%20of%20the%20solid%20that%20results%20when%20the%20region%20bounded%20by%20x%3D(√5)y2%20and%20the%20y-axis%20from%20y%3D-1%20to%20y%3D1%20is%20revolved%20around%20the%20y-axis&f=false

Answer 3

am i correct in my assumption for the equation that you typed on yahoo answers?

Answer 4

yes :)

Answer 5

ok
this is whats known as the washer method. it is called the washer method because as the object revolves around the y-axis in this case, a hole is made in the middle.
our formula will be as follows

Answer 6

\[\pi \int\limits_{c}^{d}(f(y))^{2} dy\]

Answer 7

then you just integrate, plug in your interval and tha tis your answer

Answer 8

what is the radius of the hole? I suppose that the radius of the "disc" (using the disc method, not shell) would be the original equation ([x=\sqrt{5} y ^{2}\]), right?

Answer 9

but what about the hole? or the radius is 1? I am really confused :S

Answer 10

the radius of the hole made by the revolution is accounted for in the formula

Answer 11

understand?

Answer 12

ok.
but then what would be the radius of the other part, is it 1, because it is from y=1 to y=-1.

so then the formula would be πS(1 to -1) (R-r)^2 dy
R=?
and r=\[x=\sqrt{5}y ^{2}\]