Question

Find the volume of the given solid. under the surface z=8xy and above the triangle with vertices (1,1),(4,1), and (1,2)

Answer 1

To find the volume of the solid under the surface z=8xy and above the triangle, set up a double integral with appropriate limits, integrate with respect to y, then x, and evaluate.

Step-by-step explanation:

To find the volume under the surface z=8xy and above the triangle with vertices (1,1), (4,1), and (1,2), we would integrate the function z=8xy over the triangular region defined by those points. This problem involves double integration since we are dealing with a function of two variables x and y.

The base of the solid is the triangular region in the xy-plane. We first need to find the limits of integration for both x and y, by analyzing the vertices of the triangle. The horizontal limits for x go from 1 to 4 (for the base of the triangle from (1,1) to (4,1)), and for each x, y takes on values from the line connecting (1,1) to (1,2) linearly up to the line connecting (1,1) to (4,1). The equation for the hypotenuse is y=2-x.

Now, we integrate the function 8xy with respect to y from 1 to 2-x, then integrate that with respect to x from 1 to 4. The integration will give us the volume of the desired solid.

Here's the step-by-step double integration:

1. Set up the double integral
   `\( \int_(1)^(4)\int_(1)^(2-x) 8xy \, dy \, dx \).`
2. Integrate with respect to
   `y: \( \int_(1)^(4) [4xy^2]_(1)^(2-x) \, dx \).`
3. Evaluate the integral at y=1 and y=2-x and subtract:
   `\( \int_(1)^(4) 4x[(2-x)^2-1^2] \, dx \).`
4. Finally, integrate with respect to x and evaluate to obtain the volume.

Answer 2

The volume of the given solid under the surface
`\(z = 8xy\)` and above the triangle with vertices
`\((1,1)\), \((4,1)\),` and
`\((1,2)\)` is
`\((5\pi)/(2)\)` cubic units.

To find the volume of the solid under the surface, you can set up a triple integral using cylindrical coordinates. Here's the step-by-step process:

Step 1: Determine the limits of integration for
`\(r\)`,
`\(\theta\)`, and
`\(z\)`.

1.
`\(r\) (radius)` varies from 0 to the distance from the origin to the farthest point on the triangle, which is
`\(√(3^2 + 1^2) = √(10)\)`.

2.
`\(\theta\) (angle)` can range from 0 to
`\((\pi)/(2)\)` since we are working with the first quadrant.

3.
`\(z\)` will vary from 0 to the height of the region, which is
`\(2 - 1 = 1\)`.

Step 2: Set up the triple integral for the volume:

`\[V = \iiint_E dV\]`

Where
`\(E\)` is the region in cylindrical coordinates:

`\[E = \{(r, \theta, z) : 0 \leq r \leq √(10), 0 \leq \theta \leq (\pi)/(2), 0 \leq z \leq 1\}\]`

So, the integral becomes:

`\[V = \int_0^{(\pi)/(2)} \int_0^(√(10)) \int_0^1 r \, dz \, dr \, d\theta\]`

Step 3: Evaluate the integral.

First, integrate with respect to
`\(z\)`:

`\[\int_0^1 r \, dz = r \cdot z \bigg|_0^1 = r\]`

Now, we have:

`\[V = \int_0^{(\pi)/(2)} \int_0^(√(10)) r \, dr \, d\theta\]`

Next, integrate with respect to
`\(r\)`:

`\[\int_0^(√(10)) r \, dr = (1)/(2)r^2 \bigg|_0^(√(10)) = (1)/(2)(10) - (1)/(2)(0) = 5\]`

Now, we have:

`\[V = \int_0^{(\pi)/(2)} 5 \, d\theta\]`

Finally, integrate with respect to
`\(\theta\)`:

`\[\int_0^{(\pi)/(2)} 5 \, d\theta = 5\theta \bigg|_0^{(\pi)/(2)} = 5\left((\pi)/(2)\right) - 5(0) = (5\pi)/(2)\]`

So, The answer is
`\((5\pi)/(2)\)` cubic units.