Evaluate I = ∫01 ∫12 (x²y + 4) dy dx over the rectangle 0 ≤ x ≤ 1, 1 ≤ y ≤ 2.
Multivariable calculus worksheet with answers
1.Easy
f = x²y + 40 ≤ x ≤ 11 ≤ y ≤ 2
I =
2.Easy
Evaluate I = ∫24 ∫25 (x·e−y) dy dx over the rectangle 2 ≤ x ≤ 4, 2 ≤ y ≤ 5.
f = x·e−y2 ≤ x ≤ 42 ≤ y ≤ 5
I =
3.Medium
z = 3x²y² − 3x² − 4y² + 3xy with x = 2t and y = 2t². Use the chain rule to find dz/dt at t = 1.
z = 3x²y² − 3x² − 4y² + 3xyx = 2ty = 2t²t = 1
dz/dt =
4.Medium
For f(x, y) = 2x³y³ − 5x² + 3y³ + 4xy, find the maximum rate of change of f at the point (−2, −2), that is |∇f|.
f(x, y) = 2x³y³ − 5x² + 3y³ + 4xyP = (−2, −2)
|∇f| =
5.Hard
The integral I = ∫00.8 ∫y0.8 1/(1 + x²)² dx dy cannot be done in this order. Reverse the order of integration and evaluate it.
f = 1/(1 + x²)²0 ≤ y ≤ x ≤ 0.8
I =
6.Hard
Evaluate I = ∬T (4x + y) dA over the triangle T bounded by y = 0, x = 1 and y = 3x.
f = 4x + y0 ≤ x ≤ 10 ≤ y ≤ 3x
I =
Answer key
- 1. I = 4.5
- 2. I = 0.7716
- 3. dz/dt = 236
- 4. |∇f| = 243.5
- 5. I = 0.1951
- 6. I = 5.5
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