jlajcin
Anyone know how to approach problem 5 part a from Exam 1?? The problem asks us to evaluate:
\int_0^3 \sqrt{9-x^2}\,dx.
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Problem 5a from Exam I
mark
I recommend that you approach this problem from a purely geometric perspective. You can do so by recognizing the graph of the integrand as a quarter circle of radius 3, since the following manipulations produce the equation of a circle of radius 3:
\begin{array}{ll}
y = \sqrt{9 - x^2} & \text{ } \\
y^2 = 9 - x^2 & \text{square both sides} \\
x^2+y^2 = 9 & \text{bring } x \text{ over}.
\end{array}
Thus, integral can be realized as the shaded area in the following figure:
The area of the circle is \pi \, r^2 = \pi \times3^2 = 9\pi. Thus, the value of the integral is one-fourth of that or 9\pi/4.
jlajcin
Thank you for the explanation, this makes more sense now.