Answer and Step-by-step explanation:
To evaluate the triple integral \(\iiint_{V} {F} d V\), we need to determine the limits of integration for each variable (x, y, z) based on the given region V.
1. Limits of integration for x:
The region V is bounded by x = 0 and x = 2. Therefore, the limits of integration for x are from 0 to 2.
2. Limits of integration for y:
The region V is bounded by y = 0 and y = 6. Thus, the limits of integration for y are from 0 to 6.
3. Limits of integration for z:
The region V is bounded by z = x^2 and z = 4. As x varies from 0 to 2, the corresponding limits for z are from 0^2 to 2^2, which simplifies to 0 to 4.
Now, we can set up the triple integral:
\(\iiint_{V} {F} d V = \int_{0}^{2} \int_{0}^{6} \int_{0}^{4} (2xz \hat{i} - x \hat{j} - y^2 \hat{k}) \,dz \,dy \,dx\)
To evaluate the integral, we can simplify the integrand by integrating each component separately:
\(\int_{0}^{2} \int_{0}^{6} \int_{0}^{4} 2xz \,dz \,dy \,dx - \int_{0}^{2} \int_{0}^{6} \int_{0}^{4} x \,dz \,dy \,dx - \int_{0}^{2} \int_{0}^{6} \int_{0}^{4} y^2 \,dz \,dy \,dx\)
Integrating each component:
1. \(\int_{0}^{4} 2xz \,dz = xz^2 \Big|_0^4 = 8x\)
2. \(\int_{0}^{4} x \,dz = xz \Big|_0^4 = 4x\)
3. \(\int_{0}^{4} y^2 \,dz = y^2z \Big|_0^4 = 4y^2\)
Substituting the results back into the triple integral:
\(\int_{0}^{2} \int_{0}^{6} \int_{0}^{4} (8x - 4x - 4y^2) \,dz \,dy \,dx\)
Simplifying further:
\(\int_{0}^{2} \int_{0}^{6} (4x - 4y^2) \,dy \,dx\)
Integrating with respect to y:
\(\int_{0}^{2} (4xy - \frac{4}{3}y^3) \Big|_0^6 \,dx = \int_{0}^{2} (24x - 144) \,dx\)
Integrating with respect to x:
\((12x^2 - 144x) \Big|_0^2 = 12(2^2) - 144(2) - (12(0^2) - 144(0)) = -240\)
Therefore, the evaluated value of the triple integral \(\iiint_{V} {F} d V\) is -240.