How do you evaluate the definite integral $\int_{-3}^{3} \sqrt{9-x^2}\,dx$ by interpreting it geometrically as the area of a semicircle?

Conditions

• The integrand is $\sqrt{9-x^2}$ and the limits of integration are -3 and 3.
• The square root denotes the principal (nonnegative) root, restricting the graph to $y \ge 0$.
• The function is continuous and nonnegative on the closed interval $[-3, 3]$, ensuring the definite integral equals the ordinary area under the curve.

Reasoning, step by step

1. Identify the integrand as the function $y = \sqrt{9-x^2}$.
2. Square both sides to get $y^2 = 9 - x^2$, and rearrange to the standard circle equation $x^2 + y^2 = 9$.
3. Recognize that $x^2 + y^2 = 9$ describes a full circle centered at the origin with radius $r = 3$.
4. Apply the principal square root condition $y \ge 0$ to deduce that the graph is strictly the upper semicircle.
5. Interpret the definite integral $\int_{-3}^{3} \sqrt{9-x^2}\,dx$ as the geometric area of the region bounded by the upper semicircle and the x-axis.
6. Calculate the area of the full circle using the formula $A = \pi r^2 = \pi(3)^2 = 9\pi$.
7. Halve the full circle's area to find the area of the upper semicircle: $\frac{9\pi}{2}$.
8. Conclude that the exact value of the definite integral is $\frac{9\pi}{2}$.

Example

The video demonstrates this by drawing the coordinate axes, plotting the pink upper semicircle from $(-3,0)$ through $(0,3)$ to $(3,0)$, and shading the region green. The presenter then writes the full circle area as $\pi 3^2 = 9\pi$ and divides by 2 to get $\frac{9\pi}{2}$, boxing it as the final answer for the integral.

Common misconceptions

• Assuming that because the algebraic manipulation leads to $x^2 + y^2 = 9$, the graph of $y = \sqrt{9-x^2}$ is the entire circle. It is strictly the upper semicircle because the principal square root yields only non-negative $y$-values.
• Believing that evaluating a definite integral always requires finding an antiderivative using calculus techniques. A recognizable graph can simplify integration geometrically.
• Assuming $\sqrt{9-x^2}$ can be evaluated for every real $x$. Over the reals, the radicand must be nonnegative, so the function is only defined on $[-3, 3]$.

Watch the explanation

Connected concepts

Explore next

• Why does $\sqrt{9-x^2}$ give only the top half of the circle instead of the whole circle?
• What is the real domain of $y=\sqrt{9-x^2}$ and why?
• How can this definite integral be evaluated using geometry rather than antidifferentiation?

Related questions