Science:Math Exam Resources/Courses/MATH101/April 2008/Question 07 (a)

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MATH101 April 2008

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Question 07 (a)
Show that the area of the region inside the ellipse ${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1,$ where a and b are positive constants, equals $\displaystyle \pi ab.$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
First isolate for y.

Hint 2
Find the points of intersection of the ellipse and the x-axis.

Hint 3
Note that by symmetry we need to only consider the positive square root and double the resulting value.

Hint 4
Either interpret this result as an area or try the trig substitution $x=\sin(\theta )$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

We proceed as in the hints. First, we isolate for y to see that $y^{2}=b^{2}-{\frac {b^{2}x^{2}}{a^{2}}}={\frac {b^{2}}{a^{2}}}\left(a^{2}-x^{2}\right)$ Taking the square roots yields $y=\pm {\frac {b}{a}}{\sqrt {a^{2}-x^{2}}}$ Now, we want to know the aera fo the ellipse. First, we set $y=0$ to find the points of intersection with the x-axis. This gives ${\frac {x^{2}}{a^{2}}}-0=1$ So $x=\pm a$. Thus, we wish to integrate $y=\pm {\frac {b}{a}}{\sqrt {a^{2}-x^{2}}}$ from $-a$ to $a$. By symmetry, we need only to integrate the positive function above and then double our resulting value (pictorially, the ellipse has a symmetry that is reflection about the x-axis. So to get the full area, compute half of it then double your result). Thus, we have that $A=2\int _{-a}^{a}{\frac {b}{a}}{\sqrt {a^{2}-x^{2}}}\,dx={\frac {2b}{a}}\int _{-a}^{a}{\sqrt {a^{2}-x^{2}}}\,dx$ This last integral can be interpreted as the area of a half circle of radius $a$. This has area $\displaystyle {\frac {\pi a^{2}}{2}}$. Thus, our area is $A={\frac {2b}{a}}\int _{-a}^{a}{\sqrt {a^{2}-x^{2}}}\,dx={\frac {2b}{a}}\cdot {\frac {\pi a^{2}}{2}}=\pi ab$ completing the question.

Solution 2

We proceed as in solution one until the line $A={\frac {2b}{a}}\int _{-a}^{a}{\sqrt {a^{2}-x^{2}}}\,dx$ Now we do a trig substitution. Let $x=a\sin \theta$ so that $dx=a\cos \theta d\theta$. Changing the bounds, we see that $a=a\sin(\theta )$ so $\theta ={\tfrac {\pi }{2}}$ and $-a=a\sin(\theta )$ so $\theta =-{\tfrac {\pi }{2}}$. Then the above becomes ${\begin{aligned}A&={\frac {2b}{a}}\int _{-a}^{a}{\sqrt {a^{2}-x^{2}}}\,dx\\&={\frac {2b}{a}}\int _{-{\tfrac {\pi }{2}}}^{\tfrac {\pi }{2}}{\sqrt {a^{2}-a^{2}\sin ^{2}\theta }}a\cos \theta \,d\theta \\&=2ab\int _{-{\tfrac {\pi }{2}}}^{\tfrac {\pi }{2}}\cos ^{2}\theta \,d\theta \end{aligned}}$ Now we use the double angle formula $\cos ^{2}(\theta )={\frac {1+\cos(2\theta )}{2}}$ to see that ${\begin{aligned}A&=2ab\int _{-{\tfrac {\pi }{2}}}^{\tfrac {\pi }{2}}\cos ^{2}(\theta )\,d\theta \\&=2ab\int _{-{\tfrac {\pi }{2}}}^{\tfrac {\pi }{2}}{\frac {1+\cos(2\theta )}{2}}\,d\theta \\&=\left.ab\theta \right|_{-{\tfrac {\pi }{2}}}^{\tfrac {\pi }{2}}+ab\int _{-{\tfrac {\pi }{2}}}^{\tfrac {\pi }{2}}\cos(2\theta )\,d\theta \\&={\frac {\pi ab}{2}}+{\frac {\pi ab}{2}}+\left.{\frac {ab}{2}}\sin(2\theta )\right|_{-{\tfrac {\pi }{2}}}^{\tfrac {\pi }{2}}\\&=\pi ab+{\frac {ab}{2}}(\sin(\pi )-\sin(-\pi ))\\&=\pi ab\end{aligned}}$ completing the question.

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