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Area between curves, integrating in y practice problems

Area between curves of the form x = g(y), integrating with respect to y. 10 problems with worked solutions; in 10 of them every equation is proved by a computer algebra system.

3 worked problems

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

Find the area of the region between \( \displaystyle x = y^{2} \) and \( \displaystyle x = 4 \).
  1. These are functions of y, so slice horizontally: a strip at height y runs from the left curve to the right one.
    Reviewed
  2. \[ \left. 4 - y^{2} \right|_{\substack{ y=-2 }} = 0 \]
    The curves meet at y = -2 ...✓ Proved
  3. \[ \left. 4 - y^{2} \right|_{\substack{ y=2 }} = 0 \]
    ... and at y = 2.✓ Proved
  4. \[ \int\limits_{-2}^{2} \left(4 - y^{2}\right)\, dy = \frac{32}{3} \]
    A = ∫ (right − left) dy.✓ Proved
Answer \( \frac{32}{3} \)

✓ Nihil obstat Lines: 3 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

Find the area of the region between \( \displaystyle x = y^{2} \) and \( \displaystyle x = y + 12 \).
  1. These are functions of y, so slice horizontally: a strip at height y runs from the left curve to the right one.
    Reviewed
  2. \[ \left. - y^{2} + y + 12 \right|_{\substack{ y=-3 }} = 0 \]
    The curves meet at y = -3 ...✓ Proved
  3. \[ \left. - y^{2} + y + 12 \right|_{\substack{ y=4 }} = 0 \]
    ... and at y = 4.✓ Proved
  4. \[ \int\limits_{-3}^{4} \left(- y^{2} + y + 12\right)\, dy = \frac{343}{6} \]
    A = ∫ (right − left) dy.✓ Proved
Answer \( \frac{343}{6} \)

✓ Nihil obstat Lines: 3 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

Find the area of the region between \( \displaystyle x = y^{2} \) and \( \displaystyle x = 8 - y^{2} \).
  1. These are functions of y, so slice horizontally: a strip at height y runs from the left curve to the right one.
    Reviewed
  2. \[ \left. 8 - 2 y^{2} \right|_{\substack{ y=-2 }} = 0 \]
    The curves meet at y = -2 ...✓ Proved
  3. \[ \left. 8 - 2 y^{2} \right|_{\substack{ y=2 }} = 0 \]
    ... and at y = 2.✓ Proved
  4. \[ \int\limits_{-2}^{2} \left(8 - 2 y^{2}\right)\, dy = \frac{64}{3} \]
    A = ∫ (right − left) dy.✓ Proved
Answer \( \frac{64}{3} \)

✓ Nihil obstat Lines: 3 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

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