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1 x2+y2 and A

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1. Compute

(a) Z 1

0

Z 1 0

(x + y)dydx, (b) Z 4

0

Z 12 4

x y dydx, (c) Z 1

−1

Z 1 x2

Z 2 0

(4 + z) dzdydx,

(d) Z 1

0

Z

x

0

Z 2−2x 1−x

y dzdydx,

2. Evaluate the integrals by reversing the order of integration

(a) Z 2

0

( Z 2x

x

dy)dx, (b)

Z 2 1

( Z

2x−x2

2−x

dy)dx, (c)

Z 1 0

( Z x2

x3

dy)dx, (d)

Z e 1

( Z ln x

0

dy)dx,

(e) Z 1

−2

( Z 4

y2

dx)dy, (f)

Z 1 0

( Z 2−x2

x

q dy)dx, (g) Z 2

1

( Z y

1 y

z dx)dy.

3. Find integrals of the function over the region A:

(a) f (x, y) = x · y and A is the rectangle bounded by curves x = 0, x = a, y = 0, y = b, (b) f (x, y) = 2x + y − 1 and A is the traingle with corners A(1, 1), B(5, 3), C(5, 5),

(c) f (x, y) = sin(x + y) and A is the region bounded by y = 0, y = x, x + y = π2, (d) f (x, y) = √ 1

x2+y2 and A = { (x, y) ∈ R2, a2 < x2+ y2 ≤ b2 a > 0, b > 0}, (e) f (x, y) = x2+ y2− a2 and A = { (x, y) ∈ R2, x2+ y2 ≤ ax, a > 0 }, (f) f (x, y) = y px2+ y2 and A = { (x, y) ∈ R2, x2+ y2 < 9, x < 0 },

(g) f (x, y, z) = 1−x−y1 and A is the region bounded by x + y + z = 1, x = 0, y = 0, z = 0, (h) f (x, y, z) = (18x2+ 8y2)ez and A = {(x, y, z) ∈ R3; x42 +y92 < 1, |z| < 2},

(i) f (x, y, z) = 2x + 3y − z, and A is the rectangular box bounded by x = 0, y = 0, z = 0, z = 3, x + y = 2,

(j) f (x, y, z) = z sin(x2+ y2) and A is the region bounded by x = 0, y = 0, z = 0, z = 1, x2+ y2 = 1, (k) f (x, y, z) = ze9x2+4y22 and A = { (x, y, z) ∈ R3, x42 +v92 ≤ 1, 0 ≤ z ≤ 1},

(l) f (x, y, z) = xyz and A is the region bounded by x2+ y2+ z2= 1, x = 0, y = 0, z = 0.

4. Find the area bounded by the curves

(a) y2= x, x2 = 8y, (b) 3x2= 25y, 5y2= 9x.

(c) y = x2− 2x + 2, the tangent at (3, 5), OY -axes and OX-axes, 5. Find the volume of the solid bounded by the following surfaces

(a) x2+ y2+ z2= x, (b) x42 +y92 + z2 = 1,

(c) x = 0, x = 1, y = 2, y = 5, z = 2, z = 4, (d) x + y + z = 1, x = 0, y = 0, z = 0,

(e) x42 +y92 < 1, |z| < 2,

(f) x = 0, y = 0, z = 0, z = 1, x2+ y2− 2y = 3, (g) x2+ y2+ z2= 1, x ≥ 0, y ≤ 0, z ∈ R.

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