Express y as a function of x, ( $y = f(x)$ ), for both C1 and C2.

Application of Integrals

Expressing a Circle as y = f(x) for Integration

Official Answer

For both circles, isolating y and taking the positive (upper-half) square root gives a genuine function of x.

For C1: $x^2 + y^2 = 64$

$y^2 = 64 - x^2 \Rightarrow$ $y = \sqrt{64 - x^2}$ , defined for $-8 \le x \le 8$ (upper semicircle of the roundabout boundary)

For C2: $x^2 + y^2 = 4$

$y^2 = 4 - x^2 \Rightarrow$ $y = \sqrt{4 - x^2}$ , defined for $-2 \le x \le 2$ (upper semicircle of the pond boundary)

Why the positive root: A full circle is not a function of x (it fails the vertical line test, giving two y-values for each x), so only the upper half is taken as $y = f(x)$ ; this upper-half function is what gets integrated in the next part to find the area, using the circle's symmetry.

y as a function of xsquare rootsemicirclex²+y²=r²upper halfsingle-valued function

Marking Scheme

11 mark: both $y = \sqrt{64-x^2}$ for C1 and $y = \sqrt{4-x^2}$ for C2 stated correctly (½ mark each), with the positive square root explicitly chosen.

Hint

Isolate y from $x^2+y^2=r^2$ and take only the positive square root so that y is a genuine function of x.

Quick Oral Answer

Solving $x^2+y^2=r^2$ for y gives $y=\pm\sqrt{r^2-x^2}$ ; taking the positive square root gives the upper semicircle as a function — $y=\sqrt{64-x^2}$ for C1 and $y=\sqrt{4-x^2}$ for C2.

Analysis & Explanation

This sub-question bridges the geometric picture (Q42) and the calculus to follow (Q44) by converting the circle's implicit equation into an explicit function of x.

Concept: A circle $x^2+y^2=r^2$ is a relation, not a function, because each x (except the endpoints) corresponds to two y-values. To integrate and find area, the boundary must be expressed as a genuine function $y=f(x)$ ; taking the positive square root isolates the upper semicircle, which suffices because the circle is symmetric about the x-axis, so the total area equals 4 times the area under this function from 0 to r.

Exam trap: Writing $y=\pm\sqrt{r^2-x^2}$ as the final answer does not satisfy the question, which explicitly asks for $y=f(x)$ — a function must give a single output for each x. Marks are also lost if the radius is squared incorrectly (e.g. writing $\sqrt{8-x^2}$ instead of $\sqrt{64-x^2}$ ).

Real-world application: Converting an implicit curve into an explicit function before applying calculus is the standard first move whenever areas, volumes, or centroids of curved boundaries (dams, pipes, lenses, roundabouts) must be computed using definite integrals.

Common Mistakes

1Leaving the answer as $y = \pm\sqrt{r^2-x^2}$ , which is not a function of x (two y-values per x) — the question specifically asks for $y = f(x)$ .
2Substituting the wrong radius, e.g. writing $\sqrt{8-x^2}$ instead of $\sqrt{64-x^2}$ by forgetting to square the radius before subtracting.
3Mixing up which expression belongs to C1 and which to C2.

Interesting Facts

The 'solve for y, take the positive root' step is the standard first move in nearly every area-under-a-circle or ellipse integration problem in the CBSE syllabus, including the classic NCERT worked example on the area of the circle $x^2+y^2=a^2$ .

Graphing software such as Desmos internally plots $y=\sqrt{r^2-x^2}$ and $y=-\sqrt{r^2-x^2}$ as two separate curves and stitches them together — exactly the mathematical idea being tested here.

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Frequently Asked Questions

Why can't a full circle be written as $y=f(x)$ ?

A function gives only one output y for each input x, but a circle has two y-values (positive and negative square root) for most x-values, so it fails the vertical line test; only a semicircle (one branch) can be written as a function.

Why is the positive square root chosen here?

The positive root gives the upper semicircle, which by symmetry represents exactly one-quarter (or one-half) of the enclosed area; this is the function integrated in the next part to find the total circular area.