When finding the area between two curves, you have to find the integral over some interval [a, b] of
f(x) – g(x) or f(y) – g(y).
When finding the are between y = ${\sqrt {x+3}}$ and y = ${\sqrt {3-x}}$ over the interval [-3, 3] you have to split the integral up into two separate integrals, one from [-3, 0] and one from [0, 3].
How do you recognize this when finding the area between two curves?
Determine when to split integral when calculating area between two curves? Calculus 1
calculus
Related Solutions
You'll want to find the point of intersection, and use the $t=x$-value of this point as a bound for integrating (see note below).
Set the two equations equal to one another, and solve for the values $t$ (the x-coordinate) that solves the resulting equation: there will be one such value on your domain.
$$9\sin(t) = 10\cos(t)\implies \frac{\sin t}{\cos t} = \frac {10}{9}\implies \tan t = \frac{10}{9} \implies t = \tan^{-1} \frac{10}{9}$$
Then calculate the sum of the integrals $$\int_0^t (10\cos x - 9\sin x)\,dx + \int_t^\pi (9\sin x- 10\cos x)\,dx.$$
Note: you need to know the point of intersection to divide the integral into two parts, because from $0\leq x \lt t$, where t is the value of x at the point of intersection, $10\cos x > 9\sin x$. At $x = t$, $10\cos x = 9 \sin x$. And from $t \lt x \leq \pi$, $9\sin x > 10 \cos x$. So to measure total area, we need to ensure we choose the correct function as the "upper bound" of each region when we integrate to obtain the total area of the regions bound by the functions.
You need to solve for $b$ the equation $$\frac{16}{3}=2\int^\sqrt{b}_0(b-x^2)dx$$ Which leads to $b=4^{\frac23}$
Best Answer
To expand on Fakemistake's comment:
If $\color{red}{f(x) \ge g(x)}$ for all $x \in [a,b]$, then the area between the two curves is given by $$A = \int_a^bf(x) - g(x)\,dx$$
But your two curves do not meet that if clause. Part of the problem is linguistic. The wording here is not precise. What exactly does "area between two curves" mean in this case? One could take it to mean the area between the curves horizontally:
or vertically:
Because we are expressing these curves as functions, they probably mean the vertical version.
But the area formula holds for when one curve is always above the other. In this case, the curve on top changes at $0$. So to use that area formula, you have to break the area up into the two regions where one curve is always on top, and calculate those areas separately:
$$A = A_\text{left} + A_\text{right}\\A_\text{left} = \int_{-3}^0 \sqrt{3-x} - \sqrt{x + 3}\,dx\\A_\text{right}= \int_0^3 \sqrt{x + 3} - \sqrt{3-x}\,dx$$
Alternatively, you can express the area vertically between two curves without regard to which curve is on top by $$A = \int_a^b |f(x) - g(x)|\, dx$$ But this differs from the above only superficially. You'll still end up calculating the integral of the absolute value in the same fashion.