Find the volume of a solid generated when the region bounded by the graphs of y=sqrtx, y=0 and x=4 is revolved about x=4 using a horizontal and vertical component for each  delta y and delta x.

2nd problem is the same except it is rotated around y=2
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

1. The integral can be considered to be the sum of the volumes of discs parallel to the x axis. The radius of each disc is 4-x because the widest disc has radius 4, when x=0, reducing to zero as x approaches 4. For example, at x=1 the radius is 3. The thickness of each disc is dy so the integral is (pi)S((4-x)^2dy, where S denotes integral. y=sqrt(x) so y^2=x and the integral is (pi)S((4-y^2)^2dy) between limits for y of 0 to 2, written [0,2]. The integral in y becomes:



(pi)(32-64/3+32/5)=256(pi)/15=53.62 cu units approx.

2. This time the discs are vertical with thickness dx and radius 2-y, where x goes from 0 to 4 (y ranges from 0 to 2). The integral is (pi)S[0,4]((2-y)^2dx)=




(pi)(16-1+8)=23(pi)=72.26 cu units approx.

by Top Rated User (651k points)

Related questions

1 answer
1 answer
2 answers
asked Nov 27, 2011 in Calculus Answers by anonymous | 881 views
2 answers
asked Mar 20, 2011 in Algebra 1 Answers by anonymous | 395 views
Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
83,029 questions
87,745 answers
4,553 users