Integrate G(x,y,z)=z-x over the portion of the graph of z=x+(3y^2) that lies above the triangle in the xy-plane having vertices (0,0,0), (1,1,0), and (0,1,0).

To integrate \( G(x,y,z) = z - x \) over the portion of the surface \( z = x + 3y^2 \) that lies above the triangle in the \( xy \)-plane with vertices \((0,0)\), \((1,1)\), and \((0,1)\), follow these steps:

1. **Parameterize the Surface**:
   The surface is parameterized as \( \vec{r}(x, y) = (x, y, x + 3y^2) \).

2. **Determine the Region of Integration**:
   The triangular region in the \( xy \)-plane is bounded by:
   \[
   0 \leq y \leq 1 \quad \text{and} \quad 0 \leq x \leq y.
   \]

3. **Set Up the Surface Integral**:
   The surface integral of \( G(x,y,z) = z - x \) over the surface \( S \) is:
   \[
   \iint_S (z - x) \, dS.
   \]

4. **Calculate the Surface Element \( dS \)**:
   To convert this to a double integral over the region \( R \) in the \( xy \)-plane, we use the parameterization and the surface element \( dS \):
   \[
   dS = \left\| \vec{r}_x \times \vec{r}_y \right\| \, dx \, dy,
   \]
   where \( \vec{r}_x = (1, 0, 1) \) and \( \vec{r}_y = (0, 1, 6y) \).

5. **Calculate the Cross Product**:
   \[
   \vec{r}_x \times \vec{r}_y = \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   1 & 0 & 1 \\
   0 & 1 & 6y
   \end{vmatrix} = \mathbf{i}(0 - 6y) - \mathbf{j}(1 - 0) + \mathbf{k}(1 - 0) = (-6y, -1, 1).
   \]
   The magnitude is:
   \[
   \left\| \vec{r}_x \times \vec{r}_y \right\| = \sqrt{(-6y)^2 + (-1)^2 + 1^2} = \sqrt{36y^2 + 1 + 1} = \sqrt{36y^2 + 2}.
   \]

6. **Integral in Terms of \( x \) and \( y \)**:
   Substitute \( z = x + 3y^2 \) into \( G(x,y,z) = z - x \):
   \[
   G(x,y,z) = x + 3y^2 - x = 3y^2.
   \]
   Thus, the integral becomes:
   \[
   \iint_R 3y^2 \sqrt{36y^2 + 2} \, dx \, dy.
   \]

7. **Evaluate the Double Integral**:
   Set up the double integral with the determined limits:
   \[
   \int_0^1 \int_0^y 3y^2 \sqrt{36y^2 + 2} \, dx \, dy.
   \]
   Since \( 3y^2 \sqrt{36y^2 + 2} \) is independent of \( x \), the \( dx \) integral just multiplies by \( y \):
   \[
   \int_0^1 \left( \int_0^y 3y^2 \sqrt{36y^2 + 2} \, dx \right) dy = \int_0^1 3y^3 \sqrt{36y^2 + 2} \, dy.
   \]

8. **Substitute and Solve the Remaining Integral**:
   Use the substitution \( u = 36y^2 + 2 \) with \( du = 72y \, dy \):
   \[
   y = \sqrt{\frac{u - 2}{36}}, \quad dy = \frac{1}{72} \sqrt{\frac{1}{36}} \, du = \frac{1}{12\sqrt{2}} \, du.
   \]

   The integral becomes:
   \[
   \int_{2}^{38} 3 \left( \frac{u - 2}{36} \right)^{3/2} \sqrt{u} \cdot \frac{1}{12\sqrt{2}} \, du.
   \]

   Simplify and solve the integral:
   \[
   \frac{3}{12\sqrt{2} \cdot 36^{3/2}} \int_{2}^{38} (u - 2)^{3/2} u^{1/2} \, du.
   \]

   Evaluate this integral using standard techniques for integration.

Would you like more details on the final integration step, or do you have any other questions?

Here are some related questions you might find interesting:

1. How do you parameterize a surface in general?
2. What are the steps to find the cross product of two vectors?
3. Can you explain how to determine the region of integration for a surface integral?
4. What are the common substitution techniques in integration?
5. How do you find the magnitude of a vector?

**Tip:** When solving integrals involving substitutions, carefully track your variable limits and conversions to ensure accurate integration bounds.

Learn how to integrate the function G(x,y,z)=z-x over the portion of the surface z=x+3y^2 that lies above a triangle in the xy-plane. This problem involves surface integrals, parameterization of surfaces, and the calculation of surface elements. Suitable for advanced undergraduate or graduate-level students.

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