Area of a Circle and an Ellipse by Integration (Class 12)

Understand Area of Circle and Ellipse by Integration

Area of Circle and Ellipse by Integration: formulas to remember

The key integral

$\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$

From the syllabus list of special integrals.

Quarter circle

${\int }_0^a\sqrt{a^2-x^2}dx=\frac{\pi a^2}{4}$

Use it to save time.

Area of a circle

$x^2+y^2=a^2:A=4{\int }_0^a\sqrt{a^2-x^2}dx=\pi a^2$

Symmetric about both axes.

Area of an ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1:A=\frac{4b}{a}{\int }_0^a\sqrt{a^2-x^2}dx=\pi ab$

Upper half: $y=\frac{b}{a}\sqrt{a^2-x^2}$.

Segment cut by x = c

$A=2{\int }_c^a\sqrt{a^2-x^2}dx$

For the circle; multiply by $\frac{b}{a}$ for the ellipse.

Useful values

${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6},{\sin }^{-1}\frac{1}{\sqrt{2}}=\frac{\pi }{4},{\sin }^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{3},{\sin }^{-1}1=\frac{\pi }{2}$

Appear when evaluating the key integral.

How to solve Area of Circle and Ellipse by Integration questions

1. 1Write the curve in standard form and note $a$ (and $b$); sketch it with intercepts and shade the required region.
2. 2Solve for $y$ (upper half) or for $x$ (right half) depending on the strips you choose.
3. 3Use symmetry to reduce the region to one quarter or one half, and write the multiplying factor.
4. 4Set up the integral with the correct limits (for example ${\int }_c^a$ for a segment cut by $x=c$).
5. 5Evaluate with the key formula, substituting both limits carefully; simplify the ${\sin }^{-1}$ values.
6. 6Multiply by the symmetry factor and state the area in square units.

Area of Circle and Ellipse by Integration in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Tricks to remember Area of Circle and Ellipse by Integration

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Half root, half arcsine

  $\int \sqrt{a^2-x^2}=\frac{x}{2}\sqrt{\cdot }+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}$: half times x times the root, plus half $a^2$ times arcsine.

  Use it when: Recalling the key integral.

• Quarter is πa²/4

  From 0 to a, the integral is a quarter circle: $\frac{\pi a^2}{4}$.

  Use it when: Full circles and ellipses.

• Ellipse = circle × b/a

  Every ellipse integral is the circle integral times $\frac{b}{a}$.

  Use it when: Ellipse questions.

• Segment = sector − triangle

  For a chord region, geometry gives a quick check: sector area minus triangle area.

  Use it when: Checking segment answers.

Area of Circle and Ellipse by Integration: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Where students lose marks in Area of Circle and Ellipse by Integration

The slips examiners see most often, and what to write instead.

• ✗Mistake: Forgetting the factor b/a for an ellipse

  The ellipse's upper half is $\frac{b}{a}\sqrt{a^2-x^2}$, not $\sqrt{a^2-x^2}$.

  Wrong working

  $\frac{x^2}{9}+\frac{y^2}{4}=1$: Area $=4{\int }_0^3\sqrt{9-x^2}dx=9\pi$.

  Right working

  Area $=4\cdot \frac{2}{3}{\int }_0^3\sqrt{9-x^2}dx=6\pi$.

  ✓Do this: Solve the equation for y carefully before integrating.

  Usually costs: The final answer.

• ✗Mistake: Wrong symmetry factor

  A segment cut by $x=c$ has parts above and below the axis.

  Wrong working

  Segment right of $x=2$: ${\int }_2^4\sqrt{16-x^2}dx$.

  Right working

  $2{\int }_2^4\sqrt{16-x^2}dx$.

  ✓Do this: Shade the region on a sketch and count its symmetric pieces.

  Usually costs: Half of the value.

• ✗Mistake: Dropping the first term of the key integral at the lower limit

  $\frac{x}{2}\sqrt{a^2-x^2}$ is zero at $x=0$ and $x=a$ but not at other limits.

  Wrong working

  ${\left[\dots \right]}_2^4=8\cdot \frac{\pi }{2}-8\cdot \frac{\pi }{6}$.

  Right working

  $=4\pi -\left(\sqrt{12}+\frac{4\pi }{3}\right)$.

  ✓Do this: Substitute both terms at both limits.

  Usually costs: 1 mark.

• ✗Mistake: Using degrees in sin⁻¹

  Areas need radian values: ${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$, not 30.

  Wrong working

  $8{\sin }^{-1}\frac{1}{2}=240$.

  Right working

  $8\cdot \frac{\pi }{6}=\frac{4\pi }{3}$.

  ✓Do this: Always write inverse trigonometric values in radians.

  Usually costs: The value.

Practice questions on Area of Circle and Ellipse by Integration

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The area enclosed by the circle $x^2+y^2=9$ is:
   (a) $3\pi$
   (b) $9\pi$
   (c) $6\pi$
   (d) $81\pi$

   Hint

   $\pi a^2$ with $a=3$.

   Answer

   (b) $9\pi$.

2. Q2 · 2 marks · Very short answerBoard level

   Find, using integration, the area of the part of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ in the first quadrant.

   Hint

   $\frac{4}{5}{\int }_0^5\sqrt{25-x^2}dx$.

   Answer

   $\frac{4}{5}\cdot \frac{25\pi }{4}=5\pi$ square units.

3. Q3 · 3 marks · Short answerBoard level

   Using integration, find the area of the ellipse $\frac{x^2}{4}+y^2=1$.

   Hint

   Here $a=2$, $b=1$.

   Answer

   $y=\frac{1}{2}\sqrt{4-x^2}$; Area $=4\cdot \frac{1}{2}{\int }_0^2\sqrt{4-x^2}dx=2\cdot \pi =2\pi$ square units.

4. Q4 · 3 marks · Short answerHarder board level

   Find the area of the region of the circle $x^2+y^2=9$ between the lines $x=0$ and $x=\frac{3}{2}$.

   Hint

   Use ${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$.

   Answer

   $2{\int }_0^{3/2}\sqrt{9-x^2}dx=2{\left[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}{\sin }^{-1}\frac{x}{3}\right]}_0^{3/2}=2\left[\frac{3}{4}\cdot \frac{3\sqrt{3}}{2}+\frac{9}{2}\cdot \frac{\pi }{6}\right]=\frac{9\sqrt{3}}{4}+\frac{3\pi }{2}$ square units.

5. Q5 · 1 mark · Assertion-reasonHarder board level

   Assertion (A): ${\int }_0^a\sqrt{a^2-x^2}dx=\frac{\pi a^2}{4}$.
   Reason (R): The integral is the area of the quarter of the circle $x^2+y^2=a^2$ in the first quadrant.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   What region lies under $y=\sqrt{a^2-x^2}$ from 0 to a?

   Answer

   (a). The region under the upper semicircle from 0 to a is the first-quadrant quarter circle, whose area is $\frac{\pi a^2}{4}$.

6. Q6 · 5 marks · Long answerHarder board level

   Using horizontal strips, find the area of the region of the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ above the line $y=1$.

   Hint

   For a given y, the ellipse runs from $x=-2\sqrt{4-y^2}$ to $x=2\sqrt{4-y^2}$.

   Answer

   Width at height y: $4\sqrt{4-y^2}$. Area $={\int }_1^{2}4\sqrt{4-y^2}dy=4{\left[\frac{y}{2}\sqrt{4-y^2}+2{\sin }^{-1}\frac{y}{2}\right]}_1^2=4\left[\pi -\frac{\sqrt{3}}{2}-\frac{\pi }{3}\right]=\frac{8\pi }{3}-2\sqrt{3}$ square units.

7. Q7 · 5 marks · Long answerHOTS / competency

   Find the area of the region in the first quadrant bounded by the x-axis, the line $y=x$ and the circle $x^2+y^2=18$.

   Hint

   The line meets the circle at $(3,3)$. Split at $x=3$.

   Answer

   Area $={\int }_0^{3}xdx+{\int }_3^{3\sqrt{2}}\sqrt{18-x^2}dx=\frac{9}{2}+{\left[\frac{x}{2}\sqrt{18-x^2}+9{\sin }^{-1}\frac{x}{3\sqrt{2}}\right]}_3^{3\sqrt{2}}=\frac{9}{2}+\left[\frac{9\pi }{2}-\left(\frac{9}{2}+\frac{9\pi }{4}\right)\right]=\frac{9\pi }{4}$ square units (an eighth of the circle).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Write $\int \sqrt{a^2-x^2}dx$.Show answer

   $\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$.

2. 2.What is ${\int }_0^a\sqrt{a^2-x^2}dx$?Show answer

   $\frac{\pi a^2}{4}$.

3. 3.Area of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$?Show answer

   $\pi ab$.

4. 4.Upper half of $\frac{x^2}{16}+\frac{y^2}{9}=1$?Show answer

   $y=\frac{3}{4}\sqrt{16-x^2}$.

5. 5.Area of the circle segment beyond $x=c$?Show answer

   $2{\int }_c^a\sqrt{a^2-x^2}dx$.

6. 6.Area of $\{x^2+y^2\le 4,x+y\ge 2\}$?Show answer

   $\pi -2$.

Area of Circle and Ellipse by Integration: frequently asked questions