Find an integral of differential form taken on $M^2$ manifold by Stokes' theorem

Question

Given manifold $$M^2 = \bigl\{(x, y, z): x^2 + y^2 + z^2 = 4, x \leq z \bigr\}$$ and normal $$n_x(-1, 0, 0) > 0.$$ I'm trying to find next integral of diff form: $\iint_{M^2}x^2\,dy \wedge dz + \cos x\, dx \wedge dz$ using Stokes' theorem. According to the theorem: $$d(x^2\,dy \wedge dz + \cos x\: dx \wedge dz = 2x \:dx\wedge dy \wedge dz) + 0.$$ By our constraints we have half of a sphere. In the normal coordinates х coord doesn't equal to z coord - thus, our dot is in our half-sphere. I took this parameterization for the sphere: $$(x, y, z) = (r \sin (u) \sin (v), r \cos (u), r \sin (u) \cos (v))$$ Since I need to calculate the normal for the point of the sphere, I've fixed the radius $r = 2$ and searched the normal to a circle, I suppose (I have doubts here). Found vector, orthogonal to two vectors, which sets the circle parametrization. So it's required vector. We substitute the coordinates into it from the condition and look at the fulfillment of the condition that at a given point it's positive. Since it's not running I need to find either new parametrization, or simply take "-" sign before the integral. I took second path with "-".
Further, from the result obtained by Stokes' theorem and the normal, we obtain that unknown integral is equal to the integral $$-\iiint_{M^3} 2x _,dx\wedge dy\wedge dz.$$ Here it turns out, I also need to subtract the integral $$\iint_{\partial^+M^3}... ,$$ though I'm not sure about that.
The value of a triple integral is $4\sqrt{2} \pi.$
Tell me, please, am I doing everything right, and is it necessary to subtract this integral? And what generally happens then under the sign of this integral, because there is $\cos (x)$ from given conditions?