Question

Find the area of the surface of revolution that is generated by revolving the curve x= (y^4)/8 + (y^-2)/4, from y=2 to y=5, about the line x=-1

Answer #1

Find the area of the surface generated by revolving the curve x
= ?square root 4y − y2, 1 ≤ y ≤ 2, about y-axis.

Find the volume generated by revolving the area in the first
quadrant bounded by the
curve y = e-x when the area is revolve about the line y
= -1 using the circular ring
method.

Compute the surface area of the surface generated by revolving
the curve ?(?)=(?,??)c(t)=(t,et) about the ?-x-axis for 0≤?≤1.

Determine the surface area of a funnel that is generated by
revolving the graph of y = f(x) = x^3 + (1/12x) on the interval
from [1, 2] about the x-axis.

1- Find the area enclosed by the given curves.
Find the area of the region in the first quadrant bounded on the
left by the y-axis, below by the line above left
by y = x + 4, and above right by y = - x 2 + 10.
2- Find the area enclosed by the given curves.
Find the area of the "triangular" region in the first quadrant that
is bounded above by the curve , below by the curve y...

Find the volume of the solid generated by revolving the region
bounded by y = sqrt(x) and the lines and about y=2 and x=0
about:
1) the x-axis.
2) the y-axis.
3) the line y=2.
4) the line x=4.

Find the volume of the solid generated by revolving the region
bounded by the given curve and lines about the x-axis. y=4 square
root x, y=4 x=0

Section 2 Problem 4:
a)Find the area of the surface obtained by rotating the
curvex=1/3((y^2)+2)^(3/2), 1<=y<=2, about the x-axis
b)Find the area of the surface generated by revolving the given
curve about the y-axis. x=sqrt(25-y^2), -4<=y<=4

Find the surface area of revolution for y = 2
√x over (2,4), about the x-axis.

Consider the curve y = e sin x for π /6 ≤ x ≤ π /3 . Set up the
integrals (without evaluating) that represent
1. The area of the surface generated by revolving the curve
about the x-axis.
2. The area of the surface generated by revolving the curve
about the y-axis.

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