Let R be the region bounded by y = 1/16 x^2  and the line y = 5 . Without a calculator find the volume whose base is bounded by the region R and the cross sections perpendicular to the x-axis are rectangles whose height is twice that of the side in the plane of the base

We are given the region \( R \) bounded by the parabola \( y = \frac{1}{16}x^2 \) and the line \( y = 5 \). The volume of the solid is formed by cross-sections perpendicular to the \( x \)-axis, which are rectangles where the height is twice the side in the plane of the base.

### Step 1: Determine the bounds for \( x \)
To find the intersection points between \( y = \frac{1}{16}x^2 \) and \( y = 5 \), set the two equations equal to each other:

\[
\frac{1}{16}x^2 = 5
\]

Multiply both sides by 16:

\[
x^2 = 80
\]

Taking the square root:

\[
x = \pm \sqrt{80} = \pm 4\sqrt{5}
\]

Thus, the region \( R \) is bounded by \( x = -4\sqrt{5} \) and \( x = 4\sqrt{5} \).

### Step 2: Cross-sectional area of rectangles
For each cross-section at a given \( x \), the base of the rectangle is the vertical distance between the parabola \( y = \frac{1}{16}x^2 \) and the line \( y = 5 \). This distance (or the side of the rectangle in the plane of the base) is:

\[
\text{side} = 5 - \frac{1}{16}x^2
\]

Since the height of the rectangle is twice the side in the base:

\[
\text{height} = 2 \times \left( 5 - \frac{1}{16}x^2 \right)
\]

Thus, the area \( A(x) \) of each cross-sectional rectangle is:

\[
A(x) = \text{side} \times \text{height} = \left( 5 - \frac{1}{16}x^2 \right) \times 2\left( 5 - \frac{1}{16}x^2 \right)
\]

\[
A(x) = 2 \left( 5 - \frac{1}{16}x^2 \right)^2
\]

### Step 3: Volume of the solid
The volume \( V \) of the solid is the integral of the cross-sectional area along the \( x \)-axis from \( x = -4\sqrt{5} \) to \( x = 4\sqrt{5} \):

\[
V = \int_{-4\sqrt{5}}^{4\sqrt{5}} A(x) \, dx = 2 \int_{-4\sqrt{5}}^{4\sqrt{5}} \left( 5 - \frac{1}{16}x^2 \right)^2 dx
\]

Since the integrand is an even function (it depends on \( x^2 \)), we can simplify the limits of integration by doubling the integral from 0 to \( 4\sqrt{5} \):

\[
V = 4 \int_0^{4\sqrt{5}} \left( 5 - \frac{1}{16}x^2 \right)^2 dx
\]

### Step 4: Expand and integrate
First, expand \( \left( 5 - \frac{1}{16}x^2 \right)^2 \):

\[
\left( 5 - \frac{1}{16}x^2 \right)^2 = 25 - 2 \times 5 \times \frac{1}{16}x^2 + \left( \frac{1}{16}x^2 \right)^2
\]

\[
= 25 - \frac{10}{16}x^2 + \frac{1}{256}x^4
\]

Now, integrate term by term:

\[
V = 4 \int_0^{4\sqrt{5}} \left( 25 - \frac{10}{16}x^2 + \frac{1}{256}x^4 \right) dx
\]

\[
= 4 \left[ 25x - \frac{10}{16} \frac{x^3}{3} + \frac{1}{256} \frac{x^5}{5} \right]_0^{4\sqrt{5}}
\]

### Step 5: Evaluate the integral at \( x = 4\sqrt{5} \)
Substitute \( x = 4\sqrt{5} \) into each term:

1. \( 25x = 25(4\sqrt{5}) = 100\sqrt{5} \)

2. \( \frac{10}{16} \frac{x^3}{3} = \frac{5}{8} \times \frac{(4\sqrt{5})^3}{3} = \frac{5}{8} \times \frac{64 \times 5^{3/2}}{3} = \frac{5 \times 320\sqrt{5}}{24} = \frac{1600\sqrt{5}}{24} = \frac{400\sqrt{5}}{6} \)

3. \( \frac{1}{256} \frac{x^5}{5} = \frac{1}{256} \times \frac{(4\sqrt{5})^5}{5} = \frac{1}{256} \times \frac{1024 \times 5^{5/2}}{5} = \frac{1}{256} \times 2048 \times 5^{3/2} = \frac{2048\sqrt{5}}{1280} \)

Now simplify and subtract.

Finally,

Let R be the region bounded by y = 1/16 x^2 and the line y = 5. Without a calculator, find the volume whose base is bounded by the region R and the cross-sections perpendicular to the x-axis are rectangles whose height is twice that of the side in the plane of the base.

Explore how to calculate the volume of a solid formed by rectangular cross-sections with a base bounded by the region between y = 1/16 x^2 and y = 5. This step-by-step solution applies advanced integral calculus, suitable for Grades 11-12 students.

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