Evaluate the double integral omega (x4 - 2y) \,dx\,dy over the domain omega, where omega is defined by -1 <= x <= 1 and -x2 <= y <= x2. Evaluate the triple integral E 2x \,dV over the region E under t

Here are the solutions for 6a and 6b.

a) To evaluate the integral $\iint_{\omega} (x^4 - 2y) \,dx\,dy$ over the given domain, we set up an iterated integral.

Step 1: Set up the integral with the given bounds.
The domain $\omega$ is defined by $-1 \leq x \leq 1$ and $-x^2 \leq y \leq x^2$.

$$\int_{-1}^{1} \int_{-x^2}^{x^2} (x^4 - 2y) \,dy\,dx$$

Step 2: Integrate with respect to $y$.

$$\int_{-x^2}^{x^2} (x^4 - 2y) \,dy = \left[ x^4y - y^2 \right]_{-x^2}^{x^2}$$ $$= (x^4(x^2) - (x^2)^2) - (x^4(-x^2) - (-x^2)^2)$$ $$= (x^6 - x^4) - (-x^6 - x^4)$$ $$= x^6 - x^4 + x^6 + x^4 = 2x^6$$

Step 3: Integrate the result with respect to $x$.

$$\int_{-1}^{1} 2x^6 \,dx = \left[ \frac{2x^7}{7} \right]_{-1}^{1}$$ $$= \frac{2(1)^7}{7} - \frac{2(-1)^7}{7}$$ $$= \frac{2}{7} - \left(-\frac{2}{7}\right) = \frac{2}{7} + \frac{2}{7} = \frac{4}{7}$$

The value of the integral is $\boxed{\frac{4}{7}}$.

b) To evaluate the triple integral $\iiint_E 2x \,dV$, we first define the region $E$. The region $E$ is under the plane $2x + 3y + z = 6$ and lies in the first octant ($x \geq 0, y \geq 0, z \geq 0$).

Step 1: Determine the bounds for $z$.
From the plane equation, $z = 6 - 2x - 3y$. Since $z \geq 0$, the bounds for $z$ are $0 \leq z \leq 6 - 2x - 3y$.

Step 2: Determine the bounds for $x$ and $y$.
Projecting the region onto the $xy$-plane (by setting $z=0$), we get $2x + 3y = 6$.
For $x \geq 0$ and $y \geq 0$:
The $y$-intercept is when $x=0 \Rightarrow 3y=6 \Rightarrow y=2$.
The $x$-intercept is when $y=0 \Rightarrow 2x=6 \Rightarrow x=3$.
So, $0 \leq x \leq 3$ and $0 \leq y \leq \frac{6-2x}{3}$.

Step 3: Set up the iterated integral.

$$\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}} \int_{0}^{6-2x-3y} 2x \,dz\,dy\,dx$$

Step 4: Integrate with respect to $z$.

$$\int_{0}^{6-2x-3y} 2x \,dz = [2xz]_{0}^{6-2x-3y}$$ $$= 2x(6 - 2x - 3y) - 2x(0) = 12x - 4x^2 - 6xy$$

Step 5: Integrate the result with respect to $y$.

$$\int_{0}^{\frac{6-2x}{3}} (12x - 4x^2 - 6xy) \,dy = \left[ 12xy - 4x^2y - 3xy^2 \right]_{0}^{\frac{6-2x}{3}}$$

Substitute $y = \frac{6-2x}{3}$:

$$= 12x\left(\frac{6-2x}{3}\right) - 4x^2\left(\frac{6-2x}{3}\right) - 3x\left(\frac{6-2x}{3}\right)^2$$ $$= 4x(6-2x) - \frac{4x^2}{3}(6-2x) - \frac{x}{3}(6-2x)^2$$ $$= (24x - 8x^2) - (8x^2 - \frac{8x^3}{3}) - \frac{x}{3}(36 - 24x + 4x^2)$$ $$$$