Area bounded by standard curves (circle, parabola, ellipse)

MATStandard Curves

Here the definite integral is applied to the classic conics: the area enclosed by a circle, an ellipse, a parabola with its latus rectum, or a segment cut off by a line. The standard technique exploits symmetry (computing the area of one quadrant or half and multiplying) together with the standard integral of $\sqrt{a^2-x^2}$ .

These problems test set-up, use of symmetry and a known integration result simultaneously.

Area of a full circle

A = 4\int_{0}^{a} \sqrt{a^2 - x^2}\,dx = \pi a^2

circle

x^2+y^2=a^2

of radius

a>0

; the factor

4

accounts for the four symmetric quadrants.

Standard root integral

\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right) + C

a>0

and

-a\le x\le a

; this evaluates circle and ellipse areas after substitution.

Area of an ellipse

A = 4\int_{0}^{a} \frac{b}{a}\sqrt{a^2 - x^2}\,dx = \pi a b

ellipse

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

with semi-axes

a,b>0

; reduces to

\pi a^2

when

a=b

Parabolic area up to the latus rectum

A = 2\int_{0}^{a} 2\sqrt{a x}\,dx = \frac{8a^2}{3}

parabola

y^2=4ax

(

a>0

) bounded by its latus rectum

x=a

; the factor

2

accounts for symmetry about the x-axis.

Use symmetry to shrink the work: the circle and ellipse are symmetric in all four quadrants, while the parabola $y^2=4ax$ is symmetric about the x-axis.
From $y^2=4ax$ the upper branch is $y=2\sqrt{ax}$ ; integrate this and double to include the region below the axis.
For the ellipse solve $y=\dfrac{b}{a}\sqrt{a^2-x^2}$ and pull the constant $\dfrac{b}{a}$ out of the integral.
Memorise the headline results $\pi a^2$ (circle), $\pi a b$ (ellipse) and $\dfrac{8a^2}{3}$ (parabola to latus rectum) as quick checks on your working.
For a circular segment cut by a vertical line $x=k$ , integrate $2\sqrt{a^2-x^2}$ between the appropriate x-limits.
When evaluating $\sin^{-1}$ at the limits, use $\sin^{-1}(1)=\dfrac{\pi}{2}$ and $\sin^{-1}(0)=0$ .
State clearly which region you are computing (smaller vs larger part) when a line divides a conic.
Keep the factor outside the integral consistent with the symmetry you invoked, or you will double-count or under-count.

Where the marks go

Forgetting the symmetry factor (writing $\int_0^a$ but not multiplying by $2$ or $4$ ), giving a quarter or half of the true area.
Misremembering the root integral, e.g. dropping the $\dfrac{a^2}{2}\sin^{-1}(x/a)$ term or the $\dfrac{1}{2}$ coefficients.
Confusing the ellipse semi-axes: for $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ one should read off $a=3,\,b=2$ giving $A=6\pi$ , not mishandle $a^2=9,\,b^2=4$ .
When a line cuts a circle, integrating over the wrong x-interval and returning the larger region when the smaller was asked (or vice versa).

How the board asks it

Numericalthe standard root integral and the headline area $\pi ab$
Using integration, find the area of the region enclosed by the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ .
Diagram / graphrough sketch then area between a curve and a line
Draw a rough sketch of the parabola $y^2=4x$ and the line $x=3$ , and hence find the area of the region bounded by them using integration.
Numericala circular segment cut by the vertical line $x=k$
Find the area of the smaller part of the circle $x^2+y^2=16$ cut off by the line $x=2$ .
Numericalthe parabolic area up to the latus rectum, $\dfrac{8a^2}{3}$
Find the area of the region bounded by the parabola $y^2=8x$ and its latus rectum.
Derive / provegeneral ellipse area from $y=\dfrac{b}{a}\sqrt{a^2-x^2}$
Using integration, derive an expression for the area enclosed by the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and hence show that it equals $\pi ab$ .

Practise this topic

PreviousArea between two curves

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.