MATH 1210-1: Quiz 10A
Solution
Please show your work and MARK your answer.
1. (5 points) Find the area of the plane region bounded by y = x2 and
y =x+2
Solution: Sketching the graph, we find that f (x) = x+2 is above g(x) = x2
around the plane region. Solving the equation f (x) = g(x), we have the
two intersections are at x = −1 and x = 2. By the area formula,
Z
Area =
Z
b
2
(f (x) − g(x)) dx =
a
((x + 2) − x2 ) dx =
−1
9
2
2. (5 points) Find the volume of the solid generated by revolving about the
x-axis the region bounded by the upper half of the ellipse 2x2 + y 2 = 2
and the x-axis.
Solution: The upper half of the ellipse is bounded by the function y =
√
f (x) = 2 − 2x2 on −1 6 x 6 1. By the formula of volume of solid of
revolution,
Z
Volume =
a
b
Z
πf 2 (x) dx =
1
π
−1
³p
2 − 2x2
´2
Z
dx =
1
¡
¢
8
π 2 − 2x2 dx = π
3
−1
MATH 1210-1: Quiz 10B
Solution
Please show your work and MARK your answer.
1. (5 points) Find the area of the plane region bounded by y = x2 and
y = −x + 2
Solution: Sketching the graph, we find that f (x) = −x + 2 is above
g(x) = x2 around the plane region. Solving the equation f (x) = g(x), we
have the two intersections are at x = −2 and x = 1. By the area formula,
Z
Area =
Z
b
1
(f (x) − g(x)) dx =
a
((−x + 2) − x2 ) dx =
−2
9
2
2. (5 points) Find the volume of the solid generated by revolving about the
x-axis the region bounded by the upper half of the ellipse 4x2 + y 2 = 4
and the x-axis.
Solution: The upper half of the ellipse is bounded by the function y =
√
f (x) = 4 − 4x2 on −1 6 x 6 1. By the formula of volume of solid of
revolution,
Z
Volume =
a
b
Z
πf 2 (x) dx =
1
π
−1
³p
4 − 4x2
´2
Z
dx =
1
¡
¢
16
π
π 4 − 4x2 dx =
3
−1