Area Bounded by Standard Curves

1. Recognise the standard curve and note its intercepts or vertex. 2. Draw the bounded region. 3. Decide whether vertical or horizontal strips give a simpler expression. 4. Convert the curve equation into y = f(x) or x = g(y). 5. Set limits from intercepts or intersection points. 6. Apply symmetry only if the shaded region repeats exactly. 7. Integrate or use a justified standard area result when allowed.

Sketch the standard curve accurately enough to show intercepts, axes of symmetry, and the bounded part. Label intercepts such as (-a, 0), (a, 0), (0, b), and mark whether the upper, lower, left or right branch is used.

For x^2/16 + y^2/9 = 1, writing the area as π(16)(9) instead of π(4)(3) confuses denominators with semi-axis lengths.

Area bounded by standard curves is found by first sketching the known curve shape, identifying the required bounded region, and then setting a definite integral with correct limits. Common curves include circles, parabolas, ellipses and straight lines.

Area bounded by standard curves is found by first sketching the known curve shape, identifying the required bounded region, and then setting a definite integral with correct limits. Common curves include circles, parabolas, ellipses and straight lines. Useful forms and conditions: Circle x^2 + y^2 = a^2 has upper branch y = √(a^2 - x^2), -a ≤ x ≤ a. Parabola y^2 = 4ax gives x = y^2/(4a), useful with dy. Ellipse x^2/a^2 + y^2/b^2 = 1 has upper branch y = b√(1 - x^2/a^2). Use symmetry only when the shaded region is symmetric about an axis. The area of the upper semicircle x^2 + y^2 = a^2 is ∫_-a^a √(a^2 - x^2) dx = πa^2/2 square units.

Standard curves often require choosing the correct branch or using symmetry. For a circle x^2 + y^2 = a^2, the upper half is y = √(a^2 - x^2). For a parabola y^2 = 4ax, horizontal strips may be easier because x can be written in terms of y. The diagram is not optional in reasoning; it decides which expression, limits, and multiplier are valid. Useful forms and conditions: Circle x^2 + y^2 = a^2 has upper branch y = √(a^2 - x^2), -a ≤ x ≤ a. Parabola y^2 = 4ax gives x = y^2/(4a), useful with dy. Ellipse x^2/a^2 + y^2/b^2 = 1 has upper branch y = b√(1 - x^2/a^2). Use symmetry only when the shaded region is symmetric about an axis. Find the area enclosed by the ellipse x^2/16 + y^2/9 = 1. The ellipse is symmetric about both axes. In the first quadrant, y = 3√(1 - x^2/16), with x from 0 to 4. Total area = 4∫_0^4 3√(1 - x^2/16) dx. This equals 4 times the first-quadrant area of an ellipse with semi-axes 4 and 3, so area = πab = π(4)(3) = 12π square units. The integral setup and symmetry justify the result. Questions commonly ask for area enclosed by a circle, ellipse, parabola with a line, or a region in one quadrant. Marks are awarded for a correct sketch, correct limits, justified symmetry, and valid integral setup. For x^2/16 + y^2/9 = 1, writing the area as π(16)(9) instead of π(4)(3) confuses denominators with semi-axis lengths.