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Surface area by double integrals practice problems

Surface area of z = f(x, y) over a region, by ∬ √(1 + f_x² + f_y²) dA. 60 problems with worked solutions; in 60 of them every equation is proved by a computer algebra system.

20 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 11.420

hard
Find the area of the part of the surface \( \displaystyle z = x^{2} + y^{2} \) that lies above the disk \( \displaystyle x^2 + y^2 \le 1 \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(x^{2} + y^{2}\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(x^{2} + y^{2}\right)\right)^{2} + 1 = 4 x^{2} + 4 y^{2} + 1 \]
    1 + f_x² + f_y².✓ Proved
  3. Over a disk, use polar coordinates: 1 + 4x² + 4y² = 1 + 4r², and dA = r dr dθ.
    Reviewed
  4. \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{1} r \sqrt{4 r^{2} + 1}\, dr\, d\theta = \frac{\pi \left(-1 + 5 \sqrt{5}\right)}{6} \]
    Integrate.✓ Proved
Answer \( \frac{\pi \left(-1 + 5 \sqrt{5}\right)}{6} \approx 5.33041 \)

✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.417

hard
Find the area of the part of the surface \( \displaystyle z = - 3 x - 3 y + 1 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 1] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(- 3 x - 3 y + 1\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(- 3 x - 3 y + 1\right)\right)^{2} + 1 = 19 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{3} \sqrt{19}\, dx\, dy = 3 \sqrt{19} \]
    Integrate.✓ Proved
Answer \( 3 \sqrt{19} \approx 13.07670 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.419

hard
Find the area of the part of the surface \( \displaystyle z = - x - 2 y + 4 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 3] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(- x - 2 y + 4\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(- x - 2 y + 4\right)\right)^{2} + 1 = 6 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{3}\int\limits_{0}^{2} \sqrt{6}\, dx\, dy = 6 \sqrt{6} \]
    Integrate.✓ Proved
Answer \( 6 \sqrt{6} \approx 14.69694 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.424

hard
Find the area of the part of the surface \( \displaystyle z = - x + 3 y \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(- x + 3 y\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(- x + 3 y\right)\right)^{2} + 1 = 11 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{2}\int\limits_{0}^{2} \sqrt{11}\, dx\, dy = 4 \sqrt{11} \]
    Integrate.✓ Proved
Answer \( 4 \sqrt{11} \approx 13.26650 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.427

hard
Find the area of the part of the surface \( \displaystyle z = - x + 2 y - 2 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 2] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(- x + 2 y - 2\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(- x + 2 y - 2\right)\right)^{2} + 1 = 6 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{2}\int\limits_{0}^{1} \sqrt{6}\, dx\, dy = 2 \sqrt{6} \]
    Integrate.✓ Proved
Answer \( 2 \sqrt{6} \approx 4.89898 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.430

hard
Find the area of the part of the surface \( \displaystyle z = 3 x - y - 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 1] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(3 x - y - 1\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(3 x - y - 1\right)\right)^{2} + 1 = 11 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{1} \sqrt{11}\, dx\, dy = \sqrt{11} \]
    Integrate.✓ Proved
Answer \( \sqrt{11} \approx 3.31662 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.431

hard
Find the area of the part of the surface \( \displaystyle z = 3 x - y + 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 1] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(3 x - y + 1\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(3 x - y + 1\right)\right)^{2} + 1 = 11 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{1} \sqrt{11}\, dx\, dy = \sqrt{11} \]
    Integrate.✓ Proved
Answer \( \sqrt{11} \approx 3.31662 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.483

hard
Find the area of the part of the surface \( \displaystyle z = 2 x + 2 y - 2 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(2 x + 2 y - 2\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(2 x + 2 y - 2\right)\right)^{2} + 1 = 9 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{2}\int\limits_{0}^{2} 3\, dx\, dy = 12 \]
    Integrate.✓ Proved
Answer \( 12 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.484

hard
Find the area of the part of the surface \( \displaystyle z = x - 2 y + 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 1] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(x - 2 y + 1\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(x - 2 y + 1\right)\right)^{2} + 1 = 6 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{1} \sqrt{6}\, dx\, dy = \sqrt{6} \]
    Integrate.✓ Proved
Answer \( \sqrt{6} \approx 2.44949 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.486

hard
Find the area of the part of the surface \( \displaystyle z = 2 x - 3 y + 3 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 1] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(2 x - 3 y + 3\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(2 x - 3 y + 3\right)\right)^{2} + 1 = 14 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{1} \sqrt{14}\, dx\, dy = \sqrt{14} \]
    Integrate.✓ Proved
Answer \( \sqrt{14} \approx 3.74166 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.487

hard
Find the area of the part of the surface \( \displaystyle z = - 2 x - y + 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 3] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(- 2 x - y + 1\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(- 2 x - y + 1\right)\right)^{2} + 1 = 6 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{3}\int\limits_{0}^{1} \sqrt{6}\, dx\, dy = 3 \sqrt{6} \]
    Integrate.✓ Proved
Answer \( 3 \sqrt{6} \approx 7.34847 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.488

hard
Find the area of the part of the surface \( \displaystyle z = x + y - 2 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(x + y - 2\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(x + y - 2\right)\right)^{2} + 1 = 3 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{2}\int\limits_{0}^{2} \sqrt{3}\, dx\, dy = 4 \sqrt{3} \]
    Integrate.✓ Proved
Answer \( 4 \sqrt{3} \approx 6.92820 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.493

hard
Find the area of the part of the surface \( \displaystyle z = - 3 x + 2 y - 2 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(- 3 x + 2 y - 2\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(- 3 x + 2 y - 2\right)\right)^{2} + 1 = 14 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{2}\int\limits_{0}^{2} \sqrt{14}\, dx\, dy = 4 \sqrt{14} \]
    Integrate.✓ Proved
Answer \( 4 \sqrt{14} \approx 14.96663 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.495

hard
Find the area of the part of the surface \( \displaystyle z = - x - y + 4 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 2] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(- x - y + 4\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(- x - y + 4\right)\right)^{2} + 1 = 3 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{2}\int\limits_{0}^{3} \sqrt{3}\, dx\, dy = 6 \sqrt{3} \]
    Integrate.✓ Proved
Answer \( 6 \sqrt{3} \approx 10.39230 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.496

hard
Find the area of the part of the surface \( \displaystyle z = - x + y - 2 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 1] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(- x + y - 2\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(- x + y - 2\right)\right)^{2} + 1 = 3 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{1} \sqrt{3}\, dx\, dy = \sqrt{3} \]
    Integrate.✓ Proved
Answer \( \sqrt{3} \approx 1.73205 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.497

hard
Find the area of the part of the surface \( \displaystyle z = x + y + 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 1] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(x + y + 1\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(x + y + 1\right)\right)^{2} + 1 = 3 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{1} \sqrt{3}\, dx\, dy = \sqrt{3} \]
    Integrate.✓ Proved
Answer \( \sqrt{3} \approx 1.73205 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.498

hard
Find the area of the part of the surface \( \displaystyle z = x - y - 2 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 1] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(x - y - 2\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(x - y - 2\right)\right)^{2} + 1 = 3 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{2} \sqrt{3}\, dx\, dy = 2 \sqrt{3} \]
    Integrate.✓ Proved
Answer \( 2 \sqrt{3} \approx 3.46410 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.499

hard
Find the area of the part of the surface \( \displaystyle z = x - y + 1 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(x - y + 1\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(x - y + 1\right)\right)^{2} + 1 = 3 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{2}\int\limits_{0}^{2} \sqrt{3}\, dx\, dy = 4 \sqrt{3} \]
    Integrate.✓ Proved
Answer \( 4 \sqrt{3} \approx 6.92820 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.500

hard
Find the area of the part of the surface \( \displaystyle z = x - 2 y - 2 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 3] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(x - 2 y - 2\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(x - 2 y - 2\right)\right)^{2} + 1 = 6 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{3}\int\limits_{0}^{3} \sqrt{6}\, dx\, dy = 9 \sqrt{6} \]
    Integrate.✓ Proved
Answer \( 9 \sqrt{6} \approx 22.04541 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 11.502

hard
Find the area of the part of the surface \( \displaystyle z = 2 x + y + 2 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 1] \).
  1. Surface area = ∬ √(1 + f_x² + f_y²) dA.
    Reviewed
  2. \[ \left(\frac{\partial}{\partial x} \left(2 x + y + 2\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(2 x + y + 2\right)\right)^{2} + 1 = 6 \]
    1 + f_x² + f_y².✓ Proved
  3. \[ \int\limits_{0}^{1}\int\limits_{0}^{1} \sqrt{6}\, dx\, dy = \sqrt{6} \]
    Integrate.✓ Proved
Answer \( \sqrt{6} \approx 2.44949 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More surface area by double integrals practice

Find the area of the part of the surface \( \displaystyle z = \frac{2 x^{\frac{3}{2}}}{3} + \frac{2 y^{\frac{3}{2}}}{3} \) that lies above the square \( \displaystyle [0, 1] \times [0, 1] \).
Problem 11.418hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = x^{2} + y^{2} \) that lies above the disk \( \displaystyle x^2 + y^2 \le 4 \).
Problem 11.421hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - x - 3 y + 3 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 3] \).
Problem 11.422hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = x^{2} + y^{2} \) that lies above the disk \( \displaystyle x^2 + y^2 \le \frac{1}{4} \).
Problem 11.423hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = 2 x + y - 1 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 1] \).
Problem 11.425hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = 2 x + 2 y \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 2] \).
Problem 11.426hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = - 3 x - 2 y + 1 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 1] \).
Problem 11.428hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = 3 x + 2 y + 2 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
Problem 11.429hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = - 2 x + 2 y + 4 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 1] \).
Problem 11.482hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = 3 x - y + 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 2] \).
Problem 11.485hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 2 x + y - 1 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 1] \).
Problem 11.489hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = 3 x - y \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
Problem 11.490hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = 3 x + 2 y - 2 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 1] \).
Problem 11.491hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = x - 2 y + 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 2] \).
Problem 11.492hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = 3 x - y - 2 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 2] \).
Problem 11.494hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = x - y - 1 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 3] \).
Problem 11.501hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 2 x + 2 y \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 1] \).
Problem 11.503hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 2 x - y + 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 2] \).
Problem 11.504hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = x + 3 y + 1 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
Problem 11.505hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 2 x - y + 2 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 2] \).
Problem 11.506hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 3 x + 2 y + 2 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 1] \).
Problem 11.527hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 2 x - 2 y - 2 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 3] \).
Problem 11.528hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = - 3 x + 3 y + 1 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 3] \).
Problem 11.529hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - x + y \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 1] \).
Problem 11.530hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 3 x - 2 y + 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 2] \).
Problem 11.531hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = x + 2 y - 1 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 1] \).
Problem 11.532hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = 2 x - 2 y - 2 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 1] \).
Problem 11.533hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 3 x + 3 y - 1 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 3] \).
Problem 11.534hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - x - 3 y - 1 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 1] \).
Problem 11.535hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 3 x - 2 y + 1 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
Problem 11.536hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = x + 2 y + 4 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 2] \).
Problem 11.567hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = 2 x - 2 y + 4 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 3] \).
Problem 11.568hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = 2 x + y + 1 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 1] \).
Problem 11.569hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - x + 2 y - 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 3] \).
Problem 11.570hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = - 3 x - 2 y - 2 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 1] \).
Problem 11.571hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = 2 x + y + 1 \) that lies above the rectangle \( \displaystyle [0, 2] \times [0, 2] \).
Problem 11.572hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = x + y + 3 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 3] \).
Problem 11.573hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = 3 x + 2 y + 3 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 1] \).
Problem 11.574hard✓ Nihil obstat
Find the area of the part of the surface \( \displaystyle z = - 3 x + 3 y + 1 \) that lies above the rectangle \( \displaystyle [0, 3] \times [0, 2] \).
Problem 11.575hard✓ Every equation proved
Find the area of the part of the surface \( \displaystyle z = x + 2 y + 4 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 1] \).
Problem 11.576hard✓ Nihil obstat