Question

A circular region bounded by the uh=nit circle X^{2}
+y^{2}= 1 is rotated about the liner x=4. Find the volume
of the solid (doughnut) generate.

Answer #1

1. The region bounded by y=x8 and y=sin(πx/2) is
rotated about the line x=−7.
Using cylindrical shells, set up an integral for the volume of the
resulting solid.
2.The region bounded by y=9/(1+x2), y=0, x=0 and x=8
is rotated about the line x=8.
Using cylindrical shells, set up an integral for the volume of the
resulting solid.

Find the volume of the solid generated by revolving the region
bounded by
x2−y2=16, x≥0, y=−4,y=4
about the line
x=0.

The region bounded by the given curves is rotated about the
specified axis. Find the volume V of the resulting solid
by any method.
x = (y −
9)2, x =
16; about y = 5
V =

The region bounded by y=2^x and y=4x-4 is rotated
about the line y=3. Find the volume of the resulting solid.

draw the solid bounded above z=9/2-x2-y2
and bounded below x+y+z=1. Find the volume of this
solid.

A volume is described as follows:
1. the base is the region bounded by x=−y2+8y+50x=-y2+8y+50 and
x=y2−24y+146x=y2-24y+146;
2. every cross section perpendicular to the y-axis is a
semi-circle.
Find the volume of this object.

A. For the region bounded by y = 4 − x2 and the x-axis, find
the volume of solid of revolution when the area is revolved
about:
(I) the x-axis,
(ii) the y-axis,
(iii) the line y = 4,
(iv) the line 3x + 2y − 10 = 0.
Use Second Theorem of Pappus.
B. Locate the centroid of the area of the region bounded by y
= 4 − x2 and the x-axis.

40) The region bounded by f(x)=−2x^2+12x+32, x=0 and y=0 is
rotated about the y-axis. Find the volume of the solid of
revolution.
Find the exact value; write answers without decimals.

Consider the region bounded by y = x2, y = 1, and the y-axis,
for x ≥ 0. Find the volume of the solid. The solid obtained by
rotating the region around the y-axis.

a) Find the volume of the region bounded by Z = (X2 +
Y2)2 and Z = 8 (Show all steps)
b) Find the surface area of the portion of the surface z =
X2 + Y2 which is inside the cylinder
X2 + Y2 = 2
c) Find the surface area of the portion of the graph Z = 6X + 8Y
which is above the triangle in the XY plane with vertices (0,0,0),
(2,0,0), (0,4,0)

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