Disk /Washer method

calculus2

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Hi everyone , I'm facing a huge problem with Disk/washer method , and I wish I can find a solution to my issues in this forum :

I need to find volume of the solid generated by revolving the region bounded bu the graph of f(x)= sqrt(x) , y= 0 , x=3 ,about :
a- the y-axis
b-x=3
c-x=6

The hardest part I found is determining R(x) : outter radius , and r(x)= inner radius : Can you help me please :?::?::?::?:

Thank you !
 
Hi everyone , I'm facing a huge problem with Disk/washer method , and I wish I can find a solution to my issues in this forum :

I need to find volume of the solid generated by revolving the region bounded bu the graph of f(x)= sqrt(x) , y= 0 , x=3 ,about :
a- the y-axis
b-x=3
c-x=6

The hardest part I found is determining R(x) : outter radius , and r(x)= inner radius : Can you help me please :?::?::?::?:

Thank you !

Start with sketching the region

f(x)= sqrt(x) , y= 0 , x=3

Tell us - what does it look like! May be up load the sketch.
 
Here you are :
In fact , I have the responses ,but when I solved it by my own it is not the same !! so I just need to know in each case what is R and r outer and inner radius Math.jpg
 
Here you are :
In fact , I have the responses ,but when I solved it by my own it is not the same !! so I just need to know in each case what is R and r outer and inner radius View attachment 3904
Tell us what you used for your R and r. This will help us see where you made your error and will allow us to more approriately direct to you how to correct it.

Thanks!
 
for example : around y axis :
I used only R(y)= y² and the area was 243/5* pi . It should be (36*pi*sqrt(3))/5
 
for example : around y axis :
I used only R(y)= y² and the area was 243/5* pi . It should be (36*pi*sqrt(3))/5
Remember, in looking at your arbitrary rectangle that you are rotating around the respective axis of rotation, R is the radius from the axis of rotation to the furthest point on the rectangle. In this case R = 3. Then, r is the distance from the axis of rotation to the nearest point on the rectangle. What would that be?

By the way, who said the answer was 36(sqrt 3)(pi)/5? I got a different answer.
 
Consider both the inner radius, r, and the interval or limits for integration.

Determine the y-values of the limits of integration. What limits are you using for the integration?

Let us know your R and r and the limits you're using for integration.

BTW, 36*sqrt(3)*pi / 5 is the correct solution.
 
I got it ! I used to use wrong limits i used [3,0] instead of using [sqrt(3),0] of 9-y^4 * dy ! I go the answer thank you
 
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