Difficulty: Intermediate | Prerequisites: Area Between Curves core theory notes

These worked examples come from class practice problems on area between curves. Each walks through the full solution: sketching the region, finding intersections, choosing vertical or horizontal slicing, and evaluating the integral.

Problem 1: Area of the Set y >= x², y <= 1, x >= 0

Setup: The region is bounded above by y = 1, below by y = x², and on the left by x = 0. Sketching this, it is the sliver in the first quadrant between the parabola and the horizontal line y = 1.

Finding bounds: Set y = x² equal to y = 1. That gives x² = 1, so x = 1 (taking the positive root since x >= 0). The interval is x in [0, 1].

Choosing orientation: Vertical slicing is straightforward here. Top function: f(x) = 1. Bottom function: g(x) = x².

Integral:

A = \int_0^1 (1 - x^2)\,dx = \left[x - \frac{x^3}{3}\right]_0^1 = 1 - \frac{1}{3} = \frac{2}{3}

Answer: 2/3

Problem 2: Area Enclosed by C1: x = y² - 6y and C2: x = 3y - y²

Setup: Both curves are given as x in terms of y, so horizontal integration is the natural choice.

Rewrite as functions of y:

• C1(y) = y² - 6y

• C2(y) = -y² + 3y

Finding intersections: Set C1(y) = C2(y):

y² - 6y = -y² + 3y, so 2y² - 9y = 0, giving y(2y - 9) = 0. That means y = 0 and y = 9/2.

Which is the right curve? Test a point between 0 and 9/2, say y = 1: C2(1) = -1 + 3 = 2, C1(1) = 1 - 6 = -5. So C2 is to the right of C1 on this interval.

Integral:

A = \int_0^{9/2} \bigl[C_2(y) - C_1(y)\bigr]\,dy = \int_0^{9/2} (-y^2 + 3y) - (y^2 - 6y)\,dy = \int_0^{9/2} (-2y^2 + 9y)\,dy

= \left[-\frac{2y^3}{3} + \frac{9y^2}{2}\right]_0^{9/2} = \frac{243}{8}

Answer: 243/8

Problem 3: Area Between sqrt(x) and x² on [0, 1]

Setup: Both functions are given in terms of x, and the interval is specified. Vertical slicing works directly.

Finding intersections: Set sqrt(x) = x². Squaring both sides gives x = x⁴, so x - x⁴ = 0, meaning x(1 - x³) = 0. That gives x = 0 and x = 1.

Which is on top? Test x = 0.5: sqrt(0.5) is roughly 0.71, and (0.5)² = 0.25. So sqrt(x) is the top function on [0, 1].

Integral:

A = \int_0^1 \bigl(\sqrt{x} - x^2\bigr)\,dx = \left[\frac{2x^{3/2}}{3} - \frac{x^3}{3}\right]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}

Answer: 1/3

Problem 4: Area of Region y <= 1, y >= x⁴

Setup: The region sits between y = x⁴ (below) and y = 1 (above). The curve y = x⁴ is symmetric about the y-axis, so the region extends from x = -1 to x = 1.

Finding bounds: Set x⁴ = 1. That gives x = 1 and x = -1.

Integral:

A = \int_{-1}^{1} (1 - x^4)\,dx = \left[x - \frac{x^5}{5}\right]_{-1}^{1}

= \left(1 - \frac{1}{5}\right) - \left(-1 + \frac{1}{5}\right) = \frac{4}{5} - \left(-\frac{4}{5}\right) = \frac{8}{5}

Answer: 8/5

Note: You could also use symmetry and compute 2 times the integral from 0 to 1 of (1 - x⁴) dx, which gives 2 times 4/5 = 8/5. Same result, slightly less algebra.

Problem 5: Prove the Area of a Unit Circle Is Pi

Setup: A circle of radius 1 centred at the origin has the equation x² + y² = 1. Solving for y gives the upper semicircle f(x) = sqrt(1 - x²) and the lower semicircle g(x) = -sqrt(1 - x²). The area of the full circle is the area between these two curves from x = -1 to x = 1.

Integral:

A = \int_{-1}^{1} \bigl[\sqrt{1 - x^2} - (-\sqrt{1 - x^2})\bigr]\,dx = 2\int_{-1}^{1} \sqrt{1 - x^2}\,dx

Evaluating: The antiderivative of sqrt(1 - x²) is (x sqrt(1 - x²))/2 + (1/2) arcsin(x). However, the x sqrt(1 - x²) term vanishes at both x = -1 and x = 1 (since sqrt(1 - 1) = 0), so the result simplifies to:

2 \cdot \frac{1}{2}\bigl[\arcsin(x)\bigr]_{-1}^{1} = \arcsin(1) - \arcsin(-1) = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi

Answer: pi

This is a classic exam problem. The key insight is recognising that the trig substitution (or the known antiderivative involving arcsin) is the tool that gets you from the integral to the answer.

Strategy Tips

• Always sketch the region first. A rough sketch tells you which function is on top, where they cross, and whether vertical or horizontal slicing is simpler.

• Find intersection points algebraically, then verify they make sense on the sketch.

• If the curves are given as x = something(y), integrate with respect to y. Rewriting as y = something(x) just to force vertical slicing usually creates more work, not less.

• Watch for symmetry. If the region is symmetric about an axis, you can integrate over half and double the result.

• Check your answer by units and magnitude. An area should be positive. If your integral gives a negative number, you probably subtracted in the wrong order.

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