Evaluate I = ∫13 ∫12 (4x²y − 3) dy dx over the rectangle 1 ≤ x ≤ 3, 1 ≤ y ≤ 2.
Multivariable calculus worksheet with answers
1.Easy
f = 4x²y − 31 ≤ x ≤ 31 ≤ y ≤ 2
I =
2.Easy
For f(x, y) = x²y − x³ − 4xy, find the partial derivative ∂f/∂x at the point (−2, −2).
f(x, y) = x²y − x³ − 4xyP = (−2, −2)
fx(−2, −2) =
3.Medium
z = 5x³y² − x³ − 5y² + 3xy with x = t − 1 and y = −t² + 1. Use the chain rule to find dz/dt at t = 2.
z = 5x³y² − x³ − 5y² + 3xyx = t − 1y = −t² + 1t = 2
dz/dt =
4.Medium
For f(x, y) = 5 sin(−x − 2y), find the maximum rate of change of f at the point (1, 1), that is |∇f|.
f(x, y) = 5 sin(−x − 2y)P = (1, 1)
|∇f| =
5.Hard
Use polar coordinates to evaluate I = ∬R √(x² + y²) dA, where R is the disc x² + y² ≤ 9.
f = √(x² + y²)0 ≤ r ≤ 30 ≤ θ ≤ 2π
I =
6.Hard
Evaluate I = ∬T (x + 3y) dA over the triangle T bounded by y = 0, x = 2 and y = 3x.
f = x + 3y0 ≤ x ≤ 20 ≤ y ≤ 3x
I =
Answer key
- 1. I = 46
- 2. fx(−2, −2) = 4
- 3. dz/dt = 111
- 4. |∇f| = 11.07
- 5. I = 56.55
- 6. I = 44
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