 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-354/2/c/solution

For the disk, the hemisphere reaches

z_{m a x, D} = R .

For half of the strip surface, invert the first-order equation and integrate from the midpoint to the boundary:

R = \int_{0}^{z_{m a x, T}} \frac{z ^{2} d z}{z _{m a x, T}^{4} - z ^{4}} = z_{m a x, T} \int_{0}^{1} \frac{u ^{2} d u}{1 - u ^{4}} .

\int_{0}^{1} \frac{u ^{2} d u}{1 - u ^{4}} = \frac{π Γ ( 3/4 )}{Γ ( 1/4 )} ≃ 0.59907.

Therefore

z_{m a x, T} = \frac{Γ ( 1/4 )}{π Γ ( 3/4 )} R ≃ 1.669 R > R = z_{m a x, D} .

The strip's holographic entropy surface lies deeper in the bulk than that of a disk with the same half-width or radius.

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