# Urgent!!! disk and washer method

• Apr 3rd 2008, 06:26 PM
jarny
Urgent!!! disk and washer method
I can't get these 4 problems. I just don't get how to do them. I had a sub today in math and my teacher had us learn the chapter ourselves and do it but i can't get these answers. Help w/ explaining would highly be appreciated!!! I don't care if you only do one or two, i just need to grasp the subject. 1. find the volume of the solid bounded by these equations about the indicated lines. equations: y=square root of (x) y=0 x=4. About the line x=6. 2. sames as above: equations y=2x^2 y=0 x=2 about the line y=8.

3. same above : y=6-x y=0 y=4 x=0 about the line x=6.
4. y=x * square root of (4-x^2) about y=0.

Please Hurry!!! thanks in advance and i will be waiting here trying to figure them out Thanks!!
• Apr 3rd 2008, 06:44 PM
Mathstud28
ok
The fourth one is just a basic one you need to use the formula V=pi*$\displaystyle \int_a^{b}R(x)^2-r(x)^2dx\$...so you see that you need to find the zeros of $\displaystyle x(4-x^2)^{\frac{1}{2}}\$ so either x=0 or $\displaystyle (4-x^2)^{\frac{1}{2}}\$=0 so x=0 or x=$\displaystyle \pm$2 by graphing you will see that you need to set up two integrals since [-2,0] the graph is below the x-axis and from [0,2] its above so what you do is apply the formula v=pi*$\displaystyle \int_a^bR(x)^2-r(x)^2dx\$ so V in this case is equal to pi*$\displaystyle \int_{-2}^{0}0-x(4-x^2)^{\frac{1}{2}}dx\$+pi*$\displaystyle \int_0^{2}x(4-x^2)^{\frac{1}{2}}-0,dx\$=well I will leave that up to you

the other ones when you revolve it around a vertical axis I'd use shell method which is V=2$\displaystyle pi\$$\displaystyle \int_a^{b}h(x)g(x)dx\$ where h(x) is the the distance from the axis to any point in the region you are revolving and g(x) is your curve
• Apr 3rd 2008, 06:47 PM
jarny
thanks but for 4. shouldn't you square the equation. i got 128 pi /15 for that
• Apr 3rd 2008, 06:54 PM
Mathstud28
Haha
Wow yes you aer right I suck so follow everything I said except it should now be $\displaystyle [x(4-x^2)^{\frac{1}{2}}]^2\$ in each integral..and that simplifies to $\displaystyle 4x^2-x^4\$...which then integrates to $\displaystyle \frac{4}{3}x^3-\frac{1}{5}x^5\$..haha sorry about that lol
• Apr 3rd 2008, 07:08 PM
jarny
i don't get your other formula with h(x)g(x) i don't think we've learned that yet. What i did was move the equation over for example if it was square root of (x) and it said revolve around x=4 i'd add four so it moved the equation back 4 and then revolved around the y for the answer.
• Apr 3rd 2008, 07:11 PM
Mathstud28
If you are going to do it
without the shell method you have to solve the equation so that it is no longer a function of x but now a function of y and use v=pi*$\displaystyle \int_c^d{R(y)^2}dy$
• Apr 3rd 2008, 07:13 PM
jarny
no i knew that but i was asking if there's an easier way
• Apr 3rd 2008, 07:16 PM
Mathstud28
Ok
no there is not withou the shell method...occasionally there are little teensy short cuts such as $\displaystyle \int_{-a}^{a}f(x),dx$=2$\displaystyle \int_0^{a}f(x).dx$ when f(x) is an even function but no...its really just that hard at your level...sorry(Crying)
• Apr 3rd 2008, 07:18 PM
jarny
so my ways the only way ...so far?
• Apr 3rd 2008, 07:21 PM
Mathstud28
yes
Unfortunately it is....you should learn an easier way...or Triple integrals (Rofl)...jk...just sit tight and do it the long way...it will come eventually
• Apr 3rd 2008, 07:23 PM
jarny
alright thanks bud, i really appreciate it. Finally finished(Hi) see ya
• Apr 3rd 2008, 07:26 PM
Mathstud28
Hah
no problem...Im always here if you need help!o and good job for finishing (Clapping)