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TMUA 2019 · Paper 1 · Question 9 of 20

TMUA 2019 Paper 1 Question 9

Differentiation and integration — Area between a curve and its reflection in y = x. Try it first; the answer and a full worked solution are below.

TMUA 2019 · Paper 1Differentiation and integrationArea between a curve and its reflection in y = x7 options
p is a positive constant.

Find the area enclosed between the curves y = p√x and x = p√y.

  1. A23 p52 − 12 p2
  2. B43 p52 − p2
  3. Cp46
  4. Dp43
  5. E23 p3 − 12 p4
  6. F43 p3 − p4
  7. G2p4
Show the answer and worked solution
answer · D
  1. A23 p52 − 12 p2
  2. B43 p52 − p2
  3. Cp46
  4. Dp43
  5. E23 p3 − 12 p4
  6. F43 p3 − p4
  7. G2p4
The two curves are reflections of each other in y = x. Squaring, they are y2 = p2 x and x2 = p2 y; substituting y = x2p2 into the first gives x4p4 = p2 x, so x = 0 or x = p2. Between those, y = p√x is the upper curve, so the area is ∫0p2(p x1/2 − x2p2)dx = [2p3x3/2 − x33p2]0p2 = 23 p4 − 13 p4 = p43.