Your thinking is correct, but the book is wrong in general.
Assuming that $t\in [a,b]$ with $a<b$, then considering $\displaystyle \int_a^b y \frac{dx}{dt}dt$, we can split into two cases:
(i) The integrand $y \dfrac{dx}{dt} > 0$, which can happen (a) if $y$ and $\dfrac{dx}{dt}$ are both positive, or (b) both are negative.
(ii) The integrand $y \dfrac{dx}{dt} < 0$, which can happen if either $y > 0$ and $\dfrac{dx}{dt} < 0$, vice-versa.
Both cases above correlate with your interpretation, which is correct.
Of course, the integrand can change sign within $[a,b]$, in which case the negative and positive areas will cancel each other out. Let's forget about that for now.
It is not true to say that $y \dfrac{dx}{dt} > 0$ corresponds to clockwise motion about the origin. This is a really important point.
Consider the parametric curve $x=t, y=t^2$ for $t\in [0,1]$. This is just the parabola $y=x^2$. Intuitively, one can see that in moving from $(0,0)$ ($t=0$) to $(1,1)$ ($t=1$) the curve moves anti-clockwise relative to the origin. Yet, computing the integral $\displaystyle \int_a^b y \frac{dx}{dt}dt = \dfrac{1}{3}$, which is positive.
More rigorously, we can convert the problem above to polar coordinates using the transformations $x=r\cos\theta, y=r\sin\theta$. We find that $r\geq0$ and $0\leq \theta < \dfrac{\pi}{2}$. It's quite easy to show that $\dfrac{d\theta}{dr} > 0$, thus the 'motion' is counter-clockwise throughout.
Compare this to $x=\cos\theta, y=\sin\theta$ for $\theta\in\left[0,\dfrac{\pi}{2}\right]$. The 'motion' is also counter-clockwise. However, if you compute the integral, you'll get the negative result $-\dfrac{\pi}{4}$.
The conclusion? Forget clockwise vs anticlockwise motion. What's important is the sign of $y \dfrac{dx}{dt}$, nothing more.
Either you change to polar coordinates or else you use cartesian ones. If it is as you wrote it then the parametrization is
$$-7\le x\le 7\;,\;\;-\sqrt{49-x^2}\le y\le\sqrt{49-x^2}$$
and thus
$$\oint_Cy\,dx=\int_{-7}^72\sqrt{49-x^2}\,dx=28\int_0^7\sqrt{1-\left(\frac x7\right)^2}=196\int_0^{\pi/2}\cos^2t\,dt=$$
$$=\left.98(t+\cos t\sin t)\right|_0^{\pi/2}=49\pi$$
Best Answer
The area of the top half of the ellipse is $$\int_{-a}^a y\ dx = -ab\int_{\pi}^{0} \sin^2(\varphi)\ d\varphi={\small{\frac{1}{2}}}ab\pi$$ When applying a substitution, you need to change the limits of integration to be consistent with the original limits.