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Therefore: r(R+r) V = 47 (r – R) (Vr2 – (x – R)² )dx +(47R,) (R-r) V = 4m[-;(r² – (x – R)*)}1 + 2=?r² R 47(- – (x - (R+r) (R-r) %3D 3 V = 27?r2 R

Therefore: r(R+r) V = 47 (r – R) (Vr2 – (x – R)² )dx +(47R,) (R-r) V = 4m[-;(r² – (x – R)*)}1 + 2=?r² R 47(- – (x - (R+r) (R-r) %3D 3 V = 27?r2 R

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Find the volumen of the solid using the cylindrical shell method The second picture is the problem solve, but I need to know how did they integrate (we cannot use integration by parts so if they use integration by parts I’ll need another way to do it)
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Transcribed Image Text:R wwwwww
The cross-section of the torus is a circle wwith radius r centered around (R, 0). The circle equation is: (r – R)² + y² = r² y? = r² – (x – R)² Therefore, the height of the semicircle above the x-aris is: y = Vr2 – (x – R)² Left boundary of the circle: ((R- r), 0) Right boundary of the circle: ((R+r),0) 2 We will find the volume of the torus (V) by adding cylindrical shells: r(R+r) (2nr) (2y)dar (R-r) V = r(R+r) V = (27.r) (2/r2 R)² ) dx - (r - V = 47 r(R+r) (r-R+R) (Vr2 - (r - R)2 ) dr r(R+r) (R+r) V = 4T (r-R) (Vr2 - (x - R)2 )dx + 47(R) (R-r) (V2 - ( – R)² ) dæ (R-r) The second integral is the area of a semicircle with radius r, centered around (R,0): r(R+r) (Vr2 - (r - R)² ) dx = (R-r) Vsemicircle = 2 Therefore: r(R+r) (r – R) (Vr2 - (x – R)² )dx +(47 R) (R-r) V = 47 R-r) V = 27?r2R
Transcribed Image Text:The cross-section of the torus is a circle wwith radius r centered around (R, 0). The circle equation is: (r – R)² + y² = r² y? = r² – (x – R)² Therefore, the height of the semicircle above the x-aris is: y = Vr2 – (x – R)² Left boundary of the circle: ((R- r), 0) Right boundary of the circle: ((R+r),0) 2 We will find the volume of the torus (V) by adding cylindrical shells: r(R+r) (2nr) (2y)dar (R-r) V = r(R+r) V = (27.r) (2/r2 R)² ) dx - (r - V = 47 r(R+r) (r-R+R) (Vr2 - (r - R)2 ) dr r(R+r) (R+r) V = 4T (r-R) (Vr2 - (x - R)2 )dx + 47(R) (R-r) (V2 - ( – R)² ) dæ (R-r) The second integral is the area of a semicircle with radius r, centered around (R,0): r(R+r) (Vr2 - (r - R)² ) dx = (R-r) Vsemicircle = 2 Therefore: r(R+r) (r – R) (Vr2 - (x – R)² )dx +(47 R) (R-r) V = 47 R-r) V = 27?r2R
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