Using the method of cylindrical shells, set up but do not evaluate an integral for the volume of the solid generated when the region R in the first quadrant bounded by the graphs of y = V16 – x², y = 0, and x = 0 is revolved about (a) the linex = 4 and (b) the line y = -4. (a) [ 2m(4 + x)/16 – x² åx V = V = 27(4 – x)(16 – x²) dx V = 2x(4 – x)/16 – x² ax 16 =[ «(4 – x)/16 –² åx (b) 16 V = 1(4 + y)\/16 – y dy V = 2л(4 + у) /16 —у? dy V = 27(4 – y)(16 – y²) dy 16 V = 2л (4 + у) /16 - у? dy
Using the method of cylindrical shells, set up but do not evaluate an integral for the volume of the solid generated when the region R in the first quadrant bounded by the graphs of y = V16 – x², y = 0, and x = 0 is revolved about (a) the linex = 4 and (b) the line y = -4. (a) [ 2m(4 + x)/16 – x² åx V = V = 27(4 – x)(16 – x²) dx V = 2x(4 – x)/16 – x² ax 16 =[ «(4 – x)/16 –² åx (b) 16 V = 1(4 + y)\/16 – y dy V = 2л(4 + у) /16 —у? dy V = 27(4 – y)(16 – y²) dy 16 V = 2л (4 + у) /16 - у? dy
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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Using the method of cylindrical shells, set up but do not evaluate an integral for the volume of the solid generated when the region R in the first quadrant bounded by the graphs of y=16−x2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√, y=0,y=16-x2, y=0, and x=0x=0 is revolved about (a) the line x=4x=4 and (b) the line y= −4.y= -4.
(a)
V=∫402π(4+x)16−x2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ⅆxV=∫042π(4+x)16-x2ⅆx
V=∫402π(4−x)(16−x2)ⅆxV=∫042π(4-x)(16-x2)ⅆx
V=∫402π(4−x)16−x2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ⅆxV=∫042π(4-x)16-x2ⅆx
V=∫160π(4−x)16−x2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ⅆxV=∫016π(4-x)16-x2ⅆx
(b)
V=∫160π(4+y)16−y2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ⅆyV=∫016π(4+y)16-y2ⅆy
V=∫402π(4+y)16−y2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ⅆyV=∫042π(4+y)16-y2ⅆy
V=∫402π(4−y)(16−y2)ⅆyV=∫042π(4-y)(16-y2)ⅆy
V=∫1602π(4+y)16−y2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ⅆy
Transcribed Image Text:Using the method of cylindrical shells, set up but do not evaluate an integral for the volume of the solid generated when the region Rin
the first quadrant bounded by the graphs of y = V16 – x², y = 0, and x = O is revolved about (a) the line x = 4 and (b) the line
y = -4.
(a)
V =
27(4 + x)V16 – x² dx
V =
27(4 – x)(16 – x²) ảx
V =
2n(4 – x)V16 – x? ax
16
V =
| n(4 – x)V16 –x² âx
(b)
16
V =
a(4 + y) /16 – y dy
V =
27(4 + y)\/16 – y² dy
V =
2n(4 — у) (16 — у?) dy
16
V =
2n(4 + y)/16 – y² dy
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