Evaluate I = ∫13 ∫01 (−xy² + 4) dy dx over the rectangle 1 ≤ x ≤ 3, 0 ≤ y ≤ 1.
Multivariable calculus worksheet with answers
1.Easy
f = −xy² + 41 ≤ x ≤ 30 ≤ y ≤ 1
I =
2.Easy
For f(x, y) = −4xy² − 3x² + 4y² − xy, find the mixed second partial derivative ∂²f/∂x∂y at (1, −1).
f(x, y) = −4xy² − 3x² + 4y² − xyP = (1, −1)
fxy(1, −1) =
3.Medium
z = 2x³e−y with x = 2t + 2 and y = −2t² + 2. Use the chain rule to find dz/dt at t = 1.
z = 2x³e−yx = 2t + 2y = −2t² + 2t = 1
dz/dt =
4.Medium
The tangent plane to z = 3xy³ + 4x² + y² + 2xy at the point where (x, y) = (−2, −1) can be written as z = px + qy + c. Find c.
f(x, y) = 3xy³ + 4x² + y² + 2xy(a, b) = (−2, −1)
c =
5.Hard
Evaluate I = ∬T (3x + y) dA over the triangle T bounded by y = 0, x = 2 and y = 3x.
f = 3x + y0 ≤ x ≤ 20 ≤ y ≤ 3x
I =
6.Hard
The marked points are the critical points of f(x, y) = x⁴ + y⁴ − 4xy. Click the saddle point.
f(x, y) = x⁴ + y⁴ − 4xy
- 1A (−1, −1)
- 2B (1, 1)
- 3C (0, 0)
Answer key
- 1. I = 6.667
- 2. fxy(1, −1) = 7
- 3. dz/dt = 704
- 4. c = −39
- 5. I = 36
- 6. C (0, 0)
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