AREA AND ARC LENGTH
169
21.28
Find the area enclosed by the curve y
2 = x
2 - x
4 .
Since y = x (1 —x)(l +x), the curve intersects the Jt-axis at x = Q, x = \, and x = —\. Since the
graph is symmetric with respect to the coordinate axes, it is as indicated in Fig. 21-23. The total area is four times
the area in the first quadrant, which is JJ *Vl -x
2 dx = -\ JJ (1 - *
2 )
I/2 D,(1 - x
2
) dx = -\ • |(1 -
x
2 )
3 '
2 ]
1
0 = -HO- 1) = 3. Hence, the total area is f.
Fig. 21-23
Fig. 21-24
21.29
Find the area of the loop of the curve y
2 = x
4 (4 + x) between x = -4 and x = 0. (See Fig. 21-24.)
By symmetry with respect to the x-axis, the required area is 2 J! 4 y dx = 2 J° 4 *
2 V4 + x dx. Let u =
V4 + x, u
2 = 4 + x, x = u
2 -4, dx = 2udu. Hence, we have 2 ft (u
2 - 4)
2 w • 2u du = 4 J" 0
2 (u
6 - 8w
4 +
16 M
2 )ciH = 4(^
7 -|M
5 +f M
3 )]^ = 4(^2 ? + ^)=^.
21.30
Find the length of the arc of the curve x = 3y
3 '
2 - 1 from y = 0 to y = 4.
The arc length L = J 0
4 \/l + (dx/dy)
2 dy, dx/dy=%y
1 '
2 , 1 + (dx/dy)
2 = 1+ 81y/4 = (4 + 8ly)/4. So,
L=|J 0
4 V4~+Wrf>'. Let w = 4 + 81v, du = 81 2^ (328 • 2V82 - 8) = 2S (82V82 - 1).
21.31
Find the length of the arc of 24xy = x
4 + 48 from x = 2 to x = 4.
Then
Hence,
So,
21.32 Find the length of the arc of y
3 = 8x
2
from x = 1 to x = 8.
So,
Hence,
21.33
Find the length of the arc of 6xy = x
4 + 3 from x = 1 to x = 2.
Then
and
Then
21.34 Find the length of the arc of 27 y2 = 4(x - 2)3 from (2,0) to (11,6V5).