Solid bounded by different region

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    Bounded Solid
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I think you need some face-to-face time with your teacher. I gather that English isn't your first language, but still, you don't seem to be understanding anything we are trying to tell you.
 
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LCKurtz said:
I think you need some face-to-face time with your teacher. I gather that English isn't your first language, but still, you don't seem to be understanding anything we are trying to tell you.
I agree completely. We are more than 30 posts into this problem, and you still don't understand what the solid looks like, despite multiple graphs and explanations from LCKurtz and me.
 
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LCKurtz said:
I think you need some face-to-face time with your teacher. I gather that English isn't your first language, but still, you don't seem to be understanding anything we are trying to tell you.
sorry , after looking it carefully ,the solid formed is the red part ? (refer to the attachment)
So , the red part represent the projection to xy plane ?
 
Attachments
  • 3D plot in Maple-style rendering with axes x from 0 to 4, y from -3 to 3, and z from 0 to 3, showing a translucent red solid, a teal surface and a yellow curved surface.
    424.PNG
    19.1 KB · Views: 447
here's my latest working . I use xy plane projection
 
Attachments
  • Screenshot of a triple integral calculator with inputs f(x,y,z)=1, order dz dx dy, z from -3 to 3, y from -2 to 2, x from sqrt(4-(y^2))+2 to sqrt(9-(y^2)), showing the definite integral equal to -19.461.
    425.PNG
    9.5 KB · Views: 455
  • Handwritten triple integral setup on grey paper: integral from -2 to 2 dy, inner integral from sqrt(4-y^2)+2 to sqrt(9-y^2), then integral from -3 to 3 dz dx dy, with notes x^2+y^2+z^2=9, xy plane z=0, and x=sqrt(4-y^2)+2 and x=sqrt(9-y^2).
    426.png
    41.5 KB · Views: 466
You have errors in your work. But you are supposed to do it in cylindrical coordinates anyway. You need ##dz,~dr,~d\theta## variables in your integrals.