Find the area of a circle of radius \( a \) using integrals in calculus.
Problem Solution
The equation of the circle shown above is given by:
\[ x^2 + y^2 = a^2 \]The circle is symmetric with respect to both the x and y axes. Therefore, we can find the area of one-quarter of the circle and multiply it by 4 to obtain the total area of the circle.
Solve the circle equation for \( y \):
\[ y = \pm \sqrt{a^2 - x^2} \]The equation of the upper semicircle (\( y \ge 0 \)) is given by:
\[ y = \sqrt{a^2 - x^2} \]Factor out \( a^2 \) inside the radicand:
\[ y = \sqrt{a^2 \left(1 - \frac{x^2}{a^2}\right)} \]Take \( a^2 \) from under the radicand:
\[ y = a \sqrt{1 - \frac{x^2}{a^2}} \]We use definite integrals to find the area of the upper right quarter of the circle:
Let us use trigonometric substitution by letting \( \sin t = \frac{x}{a} \), which gives \( x = a \sin t \) and \( dx = a \cos t \, dt \). The integral becomes:
\[ \dfrac{1}{4} \text{Area of circle} = \int_0^{\pi/2} a^2 \sqrt{1 - \sin^2 t} \cos t \, dt \]We now use the trigonometric identity:
\[ \sin^2 t + \cos^2 t = 1 \]which gives \( \sqrt{1 - \sin^2 t} = \cos t \) (since \( t \) varies from \( 0 \) to \( \pi/2 \)). Hence:
\[ \dfrac{1}{4} \text{Area of circle} = \int_0^{\pi/2} a^2 \cos^2 t \, dt \]Use the power-reduction identity \( \cos^2 t = \dfrac{\cos(2t) + 1}{2} \) to linearize the integrand:
\[ \dfrac{1}{4} \text{Area of circle} = \int_0^{\pi/2} a^2 \left( \dfrac{\cos(2t) + 1}{2} \right) dt \]Evaluate the integral:
\[ \dfrac{1}{4} \text{Area of circle} = \dfrac{1}{2} a^2 \left[ \dfrac{1}{2} \sin(2t) + t \right]_0^{\pi/2} \]Simplify:
\[ \dfrac{1}{4} \text{Area of circle} = \dfrac{1}{4} \pi a^2 \]The total area of the circle is obtained by multiplying the quarter-circle area by 4: