# Evaluating the definite integral $\int_0^3 \sqrt{9- x^2} \, dx$

I have been having a problem with the following definite integral:

$$\int_0^3 \sqrt{9- x^2} \, dx$$

I am only familiar with u-substitution and am positive that it can be done with only that. Any help would be appreciated.

• Are you sure on those bounds?
– Mike
May 26 '16 at 19:16
• I initially made a mistake on the bounds. It has been updated May 26 '16 at 19:23
• Hint: This can be done geometrically if you can determine the shape of the curve $y = \sqrt{9-x^2}$. May 26 '16 at 19:28

Here's a quick method: If $y = \sqrt{9-x^2}$ then $y^2 = 9-x^2$ so $x^2+y^2=9$. If you know that $x^2 + y^2 = 9$ is the equation of a circle of radius $3$ centered at $(0,0)$, that means $y^2=9-x^2$ is also an equation of that circle, as is $y = \pm\sqrt{9-x^2}$. So $y = \sqrt{9-x^2}$ (without $\text{“}\pm\text{''}$) is the top half of the circle. So the integral is just the area of one quarter of the circle. It you know the area is $\pi r^2$ and $r=3$, that tells you the answer.

(Of course, this won't work if the point is to prove that $\pi r^2$ is the area.)

Another method: \begin{align} x & = 3\sin\theta \\ dx & = 3\cos\theta\,d\theta \\ \sqrt{9-x^2} & = \sqrt{9 - 9\sin^2\theta} = \sqrt{9} \sqrt{1-\sin^2\theta} = 3\cos\theta \\[6pt] & \text{As $x$ goes from $0$ to $3$ then $\sin\theta$ goes from $0$ to $1$,} \\ & \text{so $\theta$ goes from $0$ to $\pi/2$.} \end{align} Hence the integral becomes \begin{align} & 9\int_0^{\pi/2} \cos^2\theta\,d\theta = 9\int_{\pi/2}^0 \cos^2\left(\frac\pi2 - \zeta\right) \,(-d\zeta) & & (\text{The substitution is $\theta=\frac\pi2-\zeta$}) \\[10pt] = {} & 9 \int_{\pi/2}^0 (\sin^2\zeta) \,(-d\zeta) & & \Big(\text{since $\cos\left(\frac\pi2 - \zeta\right)= \sin\zeta$}\Big) \\[10pt] = {} & 9 \int_0^{\pi/2} (\sin^2\zeta)\,d\zeta \\[10pt] = {} & 9 \int_0^{\pi/2} \sin^2\theta\,d\theta. \end{align}

So we have $$9 \int_0^{\pi/2} \cos^2\theta\,d\theta = 9 \int_0^{\pi/2} \sin^2\theta\,d\theta.$$

But on the other hand $$9 \int_0^{\pi/2} \cos^2\theta\,d\theta + 9 \int_0^{\pi/2} \sin^2\theta\,d\theta = 9 \int_0^{\pi/2} 1\,d\theta \qquad\qquad(\text{since } \cos^2\theta+\sin^2\theta =1.)$$ The latter integral comes to $9\pi/2$, so each of the two integrals separately must be half of that, or $9\pi/4$.

substitute $x$ as $3\sin \theta$, so you get $x=3\sin\theta\implies dx=-3\cos\theta \,d\theta$ and plug this in and you will get$$\int^{\pi/2}_03\cos^2\theta \,d\theta$$ and then you can use $\cos^2\theta=\dfrac{\cos2\theta+1}{2}$ and then proceed

• wouldn't you get $$3cos(\theta)$$ May 26 '16 at 19:22
• @Potato are you clear now? May 26 '16 at 19:28
• You should get $$\int_0^{\pi/2} 3\cos^2\theta \, d\theta.$$ As $x$ goes from $0$ to $3$, then $x/3=\sin\theta$ goes from $0$ to $1$, so $\theta$ goes from $0$ to $\pi/2$. And your minus sign shouldn't be there: $\dfrac d{d\theta} \sin\theta = \cos\theta\ne-\cos\theta.\qquad$ May 26 '16 at 19:38
• @MichaelHardy extremely thanks!! (just forgot about changing limits) May 26 '16 at 19:43

$$\int_{0}^{3}\sqrt{9-x^2}\space\text{d}x=$$

Substitute $x=3\sin(u)$ and $\text{d}x=3\cos(u)\space\text{d}u$.

Then $\sqrt{9-x^2}=\sqrt{9-9\sin^2(u)}=3\cos(u)$ and $u=\arcsin\left(\frac{x}{3}\right)$.

This gives a new lower bound $u=\arcsin\left(\frac{0}{3}\right)=0$ and upper bound $u=\arcsin\left(\frac{3}{3}\right)=\frac{\pi}{2}$:

$$9\int_{0}^{\frac{\pi}{2}}\cos^2(u)\space\text{d}u=$$

Use:

$$\cos^2(u)=\frac{1+\cos(2u)}{2}$$

$$\frac{9}{2}\left[\int_{0}^{\frac{\pi}{2}}\cos(2u)\space\text{d}u+\int_{0}^{\frac{\pi}{2}}1\space\text{d}u\right]=\frac{9}{2}\left[\int_{0}^{\frac{\pi}{2}}\cos(2u)\space\text{d}u+\left[u\right]_{0}^{\frac{\pi}{2}}\right]=$$

Substitute $s=2u$ and $\text{d}s=2\space\text{d}u$.

This gives a new lower bound $s=2\cdot0=0$ and upper bound $s=2\cdot\frac{\pi}{2}=\pi$:

$$\frac{9}{2}\left[\frac{1}{2}\int_{0}^{\pi}\cos(s)\space\text{d}s+\left[u\right]_{0}^{\frac{\pi}{2}}\right]=\frac{9}{2}\left[\frac{1}{2}\left[\sin(s)\right]_{0}^{\pi}+\left[u\right]_{0}^{\frac{\pi}{2}}\right]=$$ $$\frac{9}{2}\left[\frac{1}{2}\left(\sin(\pi)-\sin(0)\right)+\left(\frac{\pi}{2}-0\right)\right]=\frac{9\pi}{4}$$

Note using a possible definition of $\pi$, namely

$$\pi := 2 \cdot \int_{-1}^1 \sqrt{1 - x^2} \ dx$$ we get

$$\int_0^3 \sqrt{9- x^2} \ dx = 3\int_0^1 \sqrt{9- 9u^2} \ du = \frac{9\pi}{4}$$

• Some curricula really define $\pi$ like that? Just curious...
– Did
May 27 '16 at 6:02
• @Did Yes. For example, the Calculus textbook by Michael Spivak. May 27 '16 at 7:53
• OK, thanks. 
– Did
May 27 '16 at 8:05

Note that $y=\sqrt{9-x^2}$ is the positive half of the graph $x^2+y^2=3^2$. Since $x^2+y^2=3^2$ is a circle, $y=\sqrt{9-x^2}$ is a semicircle (with radius 3 centered at the origin). Since you are integrating from $x=0$ to $x=3$, you are only calculating half the area of the semicircle, which is one quarter the area of the full circle. The final answer is $\frac{3^2}{4}\cdot\pi$, which is equivalent to $\frac{9\pi}{4}$.

• Sorry, did not notice that @Michael Hardy already posted this solution.
– Hrhm
May 27 '16 at 4:01