Between 2 Curves
Find the area of the region enclosed by the curves:
y=2x^2-8x+10
y=(x^2/2)-2x-1
x=1
x=3
int_1^3 (2x^2-8x+10-(x^2/2-2x-1))dx
11
The base of a solid is the region enclosed by the semicircle
y=sqrt(25-x^2)
and the x-axis. Cross-sections perpendicular to the x-axis are squares.
int_-5^5 (sqrt(25-x^2))^2dx
=500/3
166.667
Find the volume of the solid that results when the region enclosed by the curves is revolved about the the y-axis.
x=2sqrt(y+4
x=0
y=1
y=2
piint_1^2 (2sqrt(y+4))^2dy
22pi
69.115
What is the chemical symbol for gold?
Au
Find the area of the region enclosed by the curves:
x=2y^2+12y+19
x=-(y^2/2)-4y-10
x=-3
x=-2
int_-3^-2 (2y^2+12y+19-(-y^2/2-4y-10))dy
=29/6
=4.833
The base of a solid is the region enclosed by
y=-x^2/9+4
and y=0. Cross-sections perpendicular to the x-axis are rectangles with heights twice that of the side in the xy-plane.
int_-6^6(-x^2/9+4)^2dx
=1024/5=204.8
Find the volume of the solid that results when the region enclosed by the curves is revolved about the the y-axis.
x=-y^2+4
x=0
y=0
y=2
piint_0^2 (-y^2+4)^2dy
256pi/15
53.617
Who directed the movie "Pulp Fiction"?
Quentin Tarantino
Find the area of the region enclosed by the curves:
y=x^2/2-3x-1/2
y=3
int_-1^7 (3-(x^2/2-3x-1/2)dx
=128/3
=42.667
The base of a solid is the region enclosed by the semicircle y=sqrt(49-x^2) and the x-axis. Cross-sections perpendicular to the x-axis are semicircles.
pi/8int_-7^7 (sqrt(49-x^2))^2dx
343pi/6
179.594
find the volume of the solid that results when the region enclosed by the curves is revolved about the the x-axis.
y=sqrtx
y=0
x=1
piint_0^1 (sqrt(x))^2dx
pi/2
1.571
What is the only continent where coffee grows naturally?
Africa
Find the area of the region enclosed by the curves:
y=-x^3/2+2x^2
y=-x^2+4x
int_0^2 (-x^2+4x-(-x^3/2+2x^2))dx+int_2^4 (-x^3/2+2x^2-(-x^2+4x))dx
4
The base of a solid is the region enclosed by y=1 and y=x^2 . Cross-sections perpendicular to the y-axis are equilateral triangles.
sqrt(3)/4 int_0^1 (sqrty+sqrty)^2dy
sqrt3/2
0.866
find the volume of the solid that results when the region enclosed by the curves is revolved about the the x-axis.
y=2/x
y=0
x=1
x=6
piint_1^6(2/x)^2dx
10pi/3
10.472
Who painted the ceiling of the Sistine Chapel?
Michelangelo
Find the area of the region enclosed by the curves:
y=-2sec^2(x)
y=2cos(x)
x=0
x=pi/4
int_0^(pi/4) (2cos(x)+2sec^2(x))dx
=2+sqrt2
=3.414
The base of a solid is the region enclosed by the circle x^2+y^2=16 . Cross-sections perpendicular to the x-axis are semicircles.
pi/8 int_-4^4(sqrt(16-x^2)+sqrt(16-x^2))^2dx
128pi/3
134.041
find the volume of the solid that results when the region enclosed by the curves is revolved about the the axis y=1.
y=-x^2+5
y=1
x=0
x=2
piint_0^2(-x^2+4)^2dx
256pi/15
53.617
In which year did World War I begin?
1914