Question

Find the area of the region bounded by the curve y²=4x and the line x=3

Hint:

Try setting the area up using horizontal strips instead: express x in terms of y for the parabola, subtract it from the fixed line x = 3 to get each strip length, and find the y limits from where the line meets the curve. Use symmetry about the x-axis to only integrate the top half and double it.

Answer 1

The region bounded by the parabola,y²=4x,and the line, x=3,is the area OACO.

The area OACO is symmetrical about x-axis.

∴Area of OACO=2(Area of OAB)

Area OACO=\(2[∫_0^3ydx]\)

=\(2∫_0^3 2\sqrt{x}dx\)

=\(4\bigg[\frac{x\frac{3}{2}}{\frac{3}{2}}\bigg]_0^3\)

=\(\frac{8}{3}[(3)^\frac{3}{2}]\)

\(=8\sqrt3\)

Therefore,the required area is \(8\sqrt3\)units.

Answer 2

Concept:

• The bounded region can also be swept out using thin horizontal strips of width $dy$, taking the strip length as (right boundary minus left boundary) at each height $y$, instead of vertical strips of width $dx$.
• Here the right boundary is the fixed line $x = 3$ and the left boundary is the parabola rewritten as $x = \dfrac{y^2}{4}$, so the strip length is simply the horizontal gap between them.

Step 1: Find the range of $y$ over which the horizontal strips run.
The line $x = 3$ meets the parabola $y^2 = 4x$ where $y^2 = 4(3) = 12$, so $y = \pm 2\sqrt{3}$. The strips run from $y = -2\sqrt{3}$ up to $y = 2\sqrt{3}$.

Step 2: Write the strip length at a general height $y$.
At height $y$, the parabola gives $x = \dfrac{y^2}{4}$ and the line gives $x = 3$, so the strip length is $3 - \dfrac{y^2}{4}$.

Step 3: Set up the area as an integral over $y$, using symmetry about the x-axis to halve the work.
$Area = 2\displaystyle\int_0^{2\sqrt{3}} \left(3 - \dfrac{y^2}{4}\right) dy$

Step 4: Integrate term by term.
$\displaystyle\int \left(3 - \dfrac{y^2}{4}\right) dy = 3y - \dfrac{y^3}{12}$
At $y = 2\sqrt{3}$: $3(2\sqrt{3}) = 6\sqrt{3}$, and $(2\sqrt{3})^3 = 8 \times 3\sqrt{3} = 24\sqrt{3}$, so $\dfrac{y^3}{12} = 2\sqrt{3}$.
This gives $6\sqrt{3} - 2\sqrt{3} = 4\sqrt{3}$.

Step 5: Multiply by the factor of $2$ from symmetry.
$Area = 2 \times 4\sqrt{3} = 8\sqrt{3}$

Final Answer: The required area is $8\sqrt{3}$ square units.