Science:Math Exam Resources/Courses/MATH101/April 2015/Question 08 (a)

MATH101 April 2015

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Question 08 (a)
Let D be the region below the graph of the curve $y={\sqrt {9-4x^{2}}}$ and above the x-axis. (a) Using an appropriate integral, find the area of the region D; simplify your answer completely.

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
Area under the graph of a function $f(x)$ and above the x-axis is given by $A=\int _{a}^{b}f(x)dx$ where $a$ and $b$ are the x-intercepts of the graph.

Hint 2
To evaluate the integral use the substitution: $x=a\sin t$ or $x=a\cos t$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First we find the integral bounds given in Hint (1), which are the points that the function crosses the x-axis. $f(x)={\sqrt {9-4x^{2}}}=0\Rightarrow 9-4x^{2}=0\Rightarrow x^{2}={\frac {9}{4}}\Rightarrow x={\frac {3}{2}},-{\frac {3}{2}}$ Now we have $A=\int _{a}^{b}f(x)dx=\int _{-{\frac {3}{2}}}^{\frac {3}{2}}{\sqrt {9-4x^{2}}}\,dx=2\int _{0}^{\frac {3}{2}}{\sqrt {4({\frac {9}{4}}-x^{2})}}\,dx=2\int _{0}^{\frac {3}{2}}2{\sqrt {{\frac {9}{4}}-x^{2}}}\,dx=4\int _{0}^{\frac {3}{2}}{\sqrt {{\frac {9}{4}}-x^{2}}}\,dx$ Note that we used the property of even functions since the integrand is even. To evaluate the integral, let $x={\frac {3}{2}}\sin t$ then $dx={\frac {3}{2}}\cos t\,dt$, then the integral bounds change to $0={\frac {3}{2}}\sin t\Rightarrow \sin t=0\Rightarrow t=0$ ${\frac {3}{2}}={\frac {3}{2}}\sin t\Rightarrow \sin t=1\Rightarrow t={\frac {\pi }{2}}$ Note that the integral is in the first quadrant, so we choose the angles in this quadrant. Now we can rewrite the integral ${\begin{aligned}A=4\int _{0}^{\frac {3}{2}}{\sqrt {{\frac {9}{4}}-x^{2}}}\,dx&=4\int _{0}^{\frac {\pi }{2}}\left({\sqrt {{\frac {9}{4}}-{\frac {9}{4}}\sin ^{2}t}}\right){\frac {3}{2}}\cos t\,dt\\&=6\int _{0}^{\frac {\pi }{2}}{\frac {3}{2}}\left({\sqrt {1-\sin ^{2}t}}\right)\cos t\,dt\\&=9\int _{0}^{\frac {\pi }{2}}\left({\sqrt {\cos ^{2}t}}\right)\cos t\,dt\\&=9\int _{0}^{\frac {\pi }{2}}|\cos t|\cos t\,dt\\&=9\int _{0}^{\frac {\pi }{2}}\cos ^{2}t\,dt\qquad \qquad {\text{ cosine is positive in the first quadrant}}\\\end{aligned}}$ Now using the identity $\cos 2t=2\cos ^{2}t-1$ we get ${\begin{aligned}A=9\int _{0}^{\frac {\pi }{2}}\cos ^{2}t\,dt&=9\int _{0}^{\frac {\pi }{2}}{\frac {1}{2}}\cos 2t+{\frac {1}{2}}dt\\&=9\left({\frac {1}{4}}\sin 2t{\bigg |}_{0}^{\frac {\pi }{2}}+{\frac {t}{2}}{\bigg |}_{0}^{\frac {\pi }{2}}\right)\\&=9\left({\frac {1}{4}}(\sin \pi -\sin 0)+{\frac {\frac {\pi }{2}}{2}}-0\right)\\\end{aligned}}$ Area: ${\color {blue}A={\frac {9\pi }{4}}}$