Science:Math Exam Resources/Courses/MATH103/April 2015/Question 03 (a)

MATH103 April 2015

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Question 03 (a)
The equation of a circle of radius $r$ centred at the origin is $x^{2}+y^{2}=r^{2}$. (a) The area A of the circle is given by $A=2\int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}dx.$ Draw a sketch, which clearly identifies the integrand, and shade the area given by $\int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}dx.$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Which part of the circle does the graph of the integrand represent?

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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. The equation for a circle centered at the origin with radius $r$, $x^{2}+y^{2}=r^{2}$, can be written as $y=\pm {\sqrt {r^{2}-x^{2}}}$. The positive part $y={\sqrt {r^{2}-x^{2}}}>0$ represents the graph for the upper half circle, while the negative part $y=-{\sqrt {r^{2}-x^{2}}}$ corresponds to the graph for the lower half circle. On the other hand, the geometric meaning of an integral $\int _{a}^{b}f(x)dx$ for the integrand satisfying $f(x)>0$ on $(a,b)$ is the area of the region enclosed by the curves $y=f(x)$, $y=0$, $x=a$, and $x=b$. Therefore, the area given by $\int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}dx$ corresponds to the upper half disk, so that the answer is as you see in the picture below. graph of upper-half circle.