Finding $\int_0^7 g(r)~dr$ from $ g(r)=\begin{cases} \sqrt{9-(r-3)^2}, & 0\le r<3 \\ r+2, & 3\le r<5 \\ \sqrt{4-(r-5)^2}, & 5 \le r \le 7\end{cases} $

Question

I have some piece-wise integration that I for some reason am not getting.

$$ g(r) = \begin{cases} \sqrt{9-(r-3)^2} && 0 \le r \lt 3 \\ r+2 && 3 \le r \lt 5 \\ \sqrt{4-(r-5)^2} && 5 \le r \le 7 \end{cases} $$

I must find the area of $\int_0^7g(r) \ dr$.

I know that this must be equivalent to:

$$ \int_0^2\sqrt{9-(r-3)^2} \ dr \ + \ \int_3^4(r+2) \ dr \ + \ \int_5^7\sqrt{4-(r-5)^2} \ dr $$

But my solution of $4.125 + 5.5 + \pi$ yields an incorrect answer?

Answer 1

I have no idea why you separated it like how you did it, this is incorrect:

$$\int_0^7 g(r)~dr= \int_0^2 \sqrt{9-(r-3)^2}~dr+\int_3^4 (r+2)~dr+\int_5^7 \sqrt{4-(r-5)^2}~dr$$

Doing your method will give this area instead:

Splitting definite integrals in two does not work like when you split summations! $$\sum_{k=0}^n f(k)\equiv \sum_{k=0}^m f(k)+\sum_{k=m+1}^n f(k)$$ $$\int_0^n f(x)~dx\not\equiv \int_0^m f(x)~dx+\int_{m+1}^n f(x)~dx$$

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In general, one can separate a definite integral into two definite ones as follows: $$\int_a^c f(x)~dx=\int_a^b f(x)~dx+\int_b^c f(x)~dx$$

Hence, you want to separate it as follows:

$$\begin{align} \int_0^7 g(r)~dr&=\int_0^3 \sqrt{9-(r-3)^2}~dr+\int_3^5 (r+2)~dr+\int_5^7 \sqrt{4-(r-5)^2}~dr \\&=\frac{9\pi}{4}+12+\pi\\&=\frac{13\pi}{4}+12\end{align}$$ Since you managed to evaluate your other integrals, I suppose you know how to evaluate these ones to $\frac{9\pi}{4}$ and $12$, so I left it as an exercise for the reader.

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Answer 2

Note: I just add this as a complement to @projectilemotion's answer.

We can still the integral into three parts as

$$\int_0^7 g(r)\; dr=\int_0^3 g(r)\; dr+\int_3^5 g(r)\; dr+\int_5^7 g(r)\; dr$$

1. For the first one, we have $y=g(r)=\sqrt{9-(r-3)^2}$, so $y^2+(r-3)^2=9$. So the region bounded by the graph of the function, the $x$-axis and the vertical lines $x=0$ and $x=3$ is a quarter of circle of center $(3,0)$ and radius. Therefore,

$$\int_0^3 g(r)\; dr=\frac{1}{4}\pi \times 3^2=\frac{9\pi}{4}$$

2. Similarly, the third integral is the area of a quarter of circle of radius $2$, so

$$\int_5^7 g(r)\; dr=\frac{1}{4}\pi \times 2^2=\pi$$

3. On $[3,5]$, the function is linear, so the region is a right trapezoid of bases $g(3)=5$ and $g(5)=7$ and height $5-3=2$, so

$$\int_3^5 g(r)\; dr=\frac{5+7}{2}\times 3=12$$

As a consequence,

$$\int_0^7 g(r)\; dr=\frac{9\pi}{4} + 12 + \pi = \frac{13\pi}{4} + 12$$