DDR 3 p.132-189
.pdfwhere limits and must be employed |
x( ) a , |
y( ) b . |
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8.1.3. Area in polar coordinates |
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Let S be the region bounded |
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1 |
α and 2 β and by the curve ρ ρ( ) , as |
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ρ ρ( ) |
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shown in Fig. 2.11. The area of S is then given by |
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S 1 |
β |
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ρ2 ( ) d . |
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α β |
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8.2. Arc length |
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Fig. 2.11 |
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arc |
The length l in rectangular coordinates of the |
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y f (x) , x [a; b] , if f |
(x) |
is continuous in this interval, then |
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dx. |
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( f (x)) |
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a |
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If the path is given parametrically, |
x x(t) , |
y y(t) |
and if x (t) , |
y (t) |
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are continuous in the interval |
[t1 ; t2 ] , then arc length is found according to |
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the following formula |
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t2 |
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l |
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dt, |
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(x (t)) |
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t1 |
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here t1 t2 .
The length l in polar coordinates of the curve ρ ρ( ) for in [ 1, 2] is equal to
2 |
2 2 d |
l |
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1 |
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8.3. Volume of a solid
8.3.1. Computing volume by parallel cross sections
The volume V of a spatial region, a “solid”, can be expressed as a definite integral of cross-section area S(x),
b
V S x dx
a
as shown in Fig. 2.12. So, to find the volume of some solid, follow these steps:
1.Choose a line to serve as an x axis.
172
2.For each plane perpendicular to that axis find the area of the cross section of the solid made by plane. Call this area S(x).
3.Determine the limits of integration, a and b, for the region.
4.Evaluate the definite integral b S x dx .
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y |
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y=f(x) |
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S(x) |
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О a |
b x |
О a |
x |
b x |
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Fig. 2.12 |
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Fig. 2.13 |
8.3.2. Computing the volume of a solid of revolution
Let y = f(x) be a continuous function such that f(x) 0 for x in the interval [a, b]. The curve y = f(x) is revolved around the x axis to form a solid of revolution (see Fig. 2.13). The volume V of a solid is then given by
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V f 2 (x)dx. |
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Let y = f1(x) |
and y = f2(x) be two continuous functions such that |
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0 f1 (x) f 2 (x) |
for x in the interval [a, b]. Let S be the region bounded by |
the curves y = f1(x) and y = f2(x) and above [a, b]. The region S is revolved around the x axis to form a solid of revolution. The volume V of a solid is then given by
b
V [ f22 (x) f12 (x)]dx.
a
8.4. Area of a surface of revolution
Let y = f(x) be a continuous function such that f(x) 0 for x in the interval [a, b] and f(x) has a continuous derivative. A formula for surface area
P if y = f(x) ( a x b ) is revolved around x axis is given by
b
P 2 f (x) 1 ( f (x))2 dx.
a
173
If a curve is given by the parametric equations x x(t) and y y(t) ,
where x(t) and y(t) have continuous derivatives and y(t) portion of the curve corresponding to t in [α, β]. Then the of revolution formed by revolving C about the x axis is
0. Let C be that area of the surface
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dt. |
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P 2 y(t) (x (t)) |
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8.5. Work
Work A of variable force F(x), if a point M is traveled from х = а to х = b, а < b, can be calculated by the integral:
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A F(x)dx. |
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8.6. The centroid |
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The centroid of a curve l, such that has an equation |
y f (x) , |
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is defined as the point (xc, yc), where |
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xdl |
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f (x) 1 ( f (x)) |
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1 ( f (x)) |
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Consider the area of the region bounded by y f (x) , and the x- axis from
x = a to x = b, as shown in Fig.2.5. The centroid of this region is defined as the point (xc, yc), where
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xf (x)dx |
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f 2 (x)dx |
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f (x)dx |
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f (x)dx |
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Typical problems |
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T8. |
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Area |
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y x 2 3x and the |
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1. Find the area of the region bounded by the curve |
line y = 4.
174
Solution. A sketch of the region is given in Fig.2.14. To determine where the curves intersect we solve the system formed by the equations y x 2 3x
and y = 4. We obtain |
y x2 3x |
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x1 1; x2 4; |
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y 4 |
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4; y2 |
4. |
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Thus the curve and the line intersect at (-1; 4) and (4; 4). The height is the y-value on the upper line minus the y-value on the lower curve:
4 |
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4 |
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S [4 (x 2 3x)]dx [4 x 2 3x]dx |
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= 16 |
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( 4 |
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20 |
5 . |
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3 4 x |
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–1 |
2. Find the area of the region bounded by the |
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Fig. 2.14 |
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curve y |
x and the lines |
y x 2 , |
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Fig. 2.15 ). |
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Solution. The line |
x 1 is passed through the intersection of the straight |
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lines y x 2 and |
y x , |
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S S1 S2 , where |
S |
is the given region, |
S1 |
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is the left part and is bounded |
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by the lines |
y x , |
x 1 and the curve |
y |
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x , S2 |
is the right part of S |
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and is bounded by the lines y x 2 , |
x 1 and |
y |
x . We have: |
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S1 ( |
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x 2 |
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S2 ( x x 2)dx |
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S S1 S2 |
5 19 |
4 . |
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175
3. |
Find the |
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of the region |
bounded |
by |
the |
two parabolas |
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x y 2 |
2y 3 |
and x 5 4 y y 2 . |
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Solution. |
Transforming |
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equations we |
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x 4 ( y 1)2 |
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(x 9) ( y 2)2 . |
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у |
х=1 |
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у=х– |
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у |
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Fig. 2.16 |
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The |
region |
is sketched |
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in |
Fig. |
2.16. The curves |
intersect |
when |
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y 2 2y 3 5 4y y 2 . Thus y 2 |
3y 4 0 , or equivalently (y + 1)(y –4) |
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= 0, from which |
y A 1 or |
yB 4 . With vertical strips three integrals are |
needed to evaluate the area.
Perhaps the use of horizontal strips can simplify our work. So, the total area is given by
4 |
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4 |
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S [5 4y y 2 |
( y 2 |
2y 3)]dy [8 6y 2 y 2 ]dy |
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8y 3y 2 |
2 y3 |
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32 48 128 ( 8 3 |
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125 . |
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4. Find the area of the less region bounded by the ellipse x 6 cos t , |
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y 4 sin t and the straight line |
y 2 3 ( y 2 3 ) . |
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Solution. The region is sketched in Fig. 2.17. The given area S is equal to |
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S 2SMNC 2(SONCD SOMCD ) . |
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At the point N xN 0 , thereby cos t 0 |
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intersection of the line and ellipse, thereby 2 |
3 4 sin t |
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The area of the rectangle OMCD is equal to:
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SOMCD OM MC 2 |
3 6 cos |
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6 3 . |
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Evaluating the area of the region ONCD : |
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SONCD 3 |
4 sin td(6 cos t) |
243 sin 2 |
tdt |
242 sin 2 tdt |
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(1 cos 2t)dt 12(t sin 2t ) |
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Hence, S 2(2 3 3 6 3) |
4 6 3 . |
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5. Find the area of the region bounded by the Bernoulli’s lemniscate |
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(x2 y 2 )2 x2 |
y 2 (Fig. 2.18). |
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Solution. Transform the given equation into a corresponding polar |
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equation. In this case the substitutions |
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x ρcos , |
y ρsin are |
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4 |
2 (cos2 sin 2 |
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cos 2 . |
Notice that there is symmetry with respect to the x axis and y axis. Consequently,
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Fig.2.17 |
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Fig. 2.18 |
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6. Find the area of the region bounded by the curves ρ 6 sin 3 and
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π |
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= 3 ( ρ 3 ). |
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Solution. |
The |
curve |
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ρ 6 sin 3 is called |
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gives one loop of the three loops |
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making |
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of |
Fig. 2.19 |
ρ 6 sin 3 . For 0 |
in [ |
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ρ 6 sin 3 is negative |
(or 0). |
This yields the lower loop in Fig. 2.19. For in [ 23 , ], is again positive,
and we obtain the upper left loop. Further choices of lead only to repetition of the loops already shown.
The second function = 3 is the circle. For |
ρ 3 the given area S is |
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shaded in Fig. 2.21 and is equal to S 6S ABCA . |
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Finding the polar coordinates of А and |
В: |
6 sin 3 3 , |
sin 3 |
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A 18 . There is В on the ray with B 6 .
As is shown the area of S ABCA SOABO SOACO , where ОАСО is a sector that is subtended by a central angle 6 . Evaluating the area of it:
SOACO 12 R 2 12 9 (p6 -18p ) = p2 .
Thereafter
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SOABO |
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36 sin 2 |
3 d 9 6 |
(1 cos 6 )d 9( sin 6 ) |
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Hence
178
S 6( 3 43 2 ) 3 9 23 .
Arc length
7. Find the arc length of the curve y =x2 for x in 0,1 .
Solution. By the given formula l b |
1 y 2 dx. |
a |
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Since y = x2, y 2x. Thus l 1 1 4x2 dx
0
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8.
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2x tdt dx |
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Find the length of one arch of the cycloid |
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x a(t sin t) , |
y a(1 cos t) . |
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Solution. Here the parameter is t and we compute xt/ and yt/ :
xt/ a 1 cos t ;yt/ a sin t.
To complete, one arch varies from 0 to 2 (see Fig. 2.20). By given
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xt/ 2 |
yt/ 2 d |
formula the length of one arch is: l |
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179
y
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О• |
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Fig. 2.20 |
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2πa |
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4 sin 2 |
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sin |
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Find the arc length of the curve ρ sin (Fig. 2.21). |
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y |
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Solution. Since 0, then 0 . Hence |
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sin2 cos2 d d . |
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Remark. It is no surprise that the graph appears to be a |
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Fig. 2.21 |
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x 2 ( y 1/ 2)2 1/ 4 . |
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Volume of a solid |
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12. Find the volume of the solid bounded by the |
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paraboloid |
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(Fig.2.25). |
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Solution. There is a region between the paraboloid |
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and cone. The volume V of it is equal to |
VVп Vк ,
Оywhere Vn is the volume of the paraboloid between the
xpoint z = 0 and the plane z 1 (they are the roots of the
Fig. 2.25 |
equation z z 2 ) and Vk is the volume of the cone |
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between them. |
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The perpendicular section to z axis are the ellipses |
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known the area of ellipse |
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is ab. |
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1 |
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V Vп Vк |
z 2 |
zdz z 2zdz 2zdz |
2z 2 dz |
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13. The region between x 2 y 2 |
1, |
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2x 2 and x 0 ( x 0 ) is |
revolved around the a) x axis (Fig. 2.26); b) y axis (Fig.2.27). Find the volume of the solids of revolution produced.
y |
у |
1 |
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1 x |
1 х |
Fig. 2.26 |
Fig. 2.27 |
Solution. Evaluating the system
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x |
3 |
x 2 |
y 2 1, |
y |
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2x 2 , |
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16x |
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x |
we get x 22 and y 22 . Thereafter
181