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To find the area of the region included between the parabolas [tex]\( y^2 = 16x \)[/tex] and [tex]\( x^2 = 16y \)[/tex], we need to follow several steps:
### Step 1: Find the Points of Intersection
First, we need to find the points at which these two parabolas intersect.
1. Rewrite the equations:
- For [tex]\( y^2 = 16x \)[/tex]: [tex]\( y = \pm \sqrt{16x} \)[/tex]
- For [tex]\( x^2 = 16y \)[/tex]: [tex]\( y = \frac{x^2}{16} \)[/tex]
We'll use the positive branch of [tex]\( y \)[/tex] for both since we're interested in the region where both functions are non-negative:
- [tex]\( y = \sqrt{16x} \)[/tex]
- [tex]\( y = \frac{x^2}{16} \)[/tex]
2. Set them equal to each other to find [tex]\( x \)[/tex]:
[tex]\[ \sqrt{16x} = \frac{x^2}{16} \][/tex]
3. Square both sides to eliminate the square root:
[tex]\[ 16x = \left( \frac{x^2}{16} \right)^2 \][/tex]
[tex]\[ 16x = \frac{x^4}{256} \][/tex]
4. Multiply through by 256 to clear the fraction:
[tex]\[ 4096x = x^4 \][/tex]
5. Rearrange and solve for [tex]\( x \)[/tex]:
[tex]\[ x^4 - 4096x = 0 \][/tex]
[tex]\[ x(x^3 - 4096) = 0 \][/tex]
[tex]\[ x = 0 \quad \text{or} \quad x^3 = 4096 \][/tex]
[tex]\[ x = 0 \quad \text{or} \quad x = \sqrt[3]{4096} = 16 \][/tex]
So, the points of intersection are [tex]\( (0, 0) \)[/tex] and [tex]\( (16, 64) \)[/tex].
### Step 2: Set Up and Evaluate the Integral
To find the area between the curves, we integrate the difference of the functions over the interval [tex]\( [0, 16] \)[/tex].
1. The upper function [tex]\( y = \sqrt{16x} \)[/tex] and the lower function [tex]\( y = \frac{x^2}{16} \)[/tex].
2. Set up the integral:
[tex]\[ \text{Area} = \int_{0}^{16} \left( \sqrt{16x} - \frac{x^2}{16} \right) \, dx \][/tex]
### Step 3: Solve the Integral
1. Simplify the integrand:
[tex]\[ \sqrt{16x} = 4\sqrt{x} \][/tex]
[tex]\[ \text{Area} = \int_{0}^{16} \left( 4\sqrt{x} - \frac{x^2}{16} \right) \, dx \][/tex]
2. Compute the definite integral:
[tex]\[ \int 4\sqrt{x} \, dx = \int 4x^{1/2} \, dx = 4 \cdot \frac{2}{3} x^{3/2} = \frac{8}{3} x^{3/2} \][/tex]
[tex]\[ \int \frac{x^2}{16} \, dx = \frac{1}{16} \int x^2 \, dx = \frac{1}{16} \cdot \frac{x^3}{3} = \frac{1}{48} x^3 \][/tex]
3. Evaluate the integrals from 0 to 16:
[tex]\[ \left[ \frac{8}{3} x^{3/2} \right]_{0}^{16} - \left[ \frac{1}{48} x^3 \right]_{0}^{16} \][/tex]
[tex]\[ \left( \frac{8}{3} (16)^{3/2} \right) - \left( \frac{1}{48} (16)^3 \right) \][/tex]
4. Calculate [tex]\( (16)^{3/2} \)[/tex] and [tex]\( (16)^3 \)[/tex]:
[tex]\[ (16)^{3/2} = \sqrt{16^3} = \sqrt{4096} = 64 \][/tex]
[tex]\[ (16)^3 = 4096 \][/tex]
5. Substitute back into the integrals:
[tex]\[ \frac{8}{3} \cdot 64 - \frac{1}{48} \cdot 4096 \][/tex]
[tex]\[ \frac{512}{3} - \frac{4096}{48} \][/tex]
[tex]\[ \frac{512}{3} - \frac{4096}{48} = \frac{512}{3} - \frac{4096}{48} \][/tex]
6. Simplify [tex]\(\frac{4096}{48}\)[/tex]:
[tex]\[ \frac{4096}{48} = \frac{1024}{12} = \frac{256}{3} \][/tex]
7. Subtract the fractions:
[tex]\[ \frac{512}{3} - \frac{256}{3} = \frac{256}{3} \][/tex]
Thus, the area of the region included between the parabolas is [tex]\(\boxed{\frac{256}{3}}\)[/tex] square units.
### Step 1: Find the Points of Intersection
First, we need to find the points at which these two parabolas intersect.
1. Rewrite the equations:
- For [tex]\( y^2 = 16x \)[/tex]: [tex]\( y = \pm \sqrt{16x} \)[/tex]
- For [tex]\( x^2 = 16y \)[/tex]: [tex]\( y = \frac{x^2}{16} \)[/tex]
We'll use the positive branch of [tex]\( y \)[/tex] for both since we're interested in the region where both functions are non-negative:
- [tex]\( y = \sqrt{16x} \)[/tex]
- [tex]\( y = \frac{x^2}{16} \)[/tex]
2. Set them equal to each other to find [tex]\( x \)[/tex]:
[tex]\[ \sqrt{16x} = \frac{x^2}{16} \][/tex]
3. Square both sides to eliminate the square root:
[tex]\[ 16x = \left( \frac{x^2}{16} \right)^2 \][/tex]
[tex]\[ 16x = \frac{x^4}{256} \][/tex]
4. Multiply through by 256 to clear the fraction:
[tex]\[ 4096x = x^4 \][/tex]
5. Rearrange and solve for [tex]\( x \)[/tex]:
[tex]\[ x^4 - 4096x = 0 \][/tex]
[tex]\[ x(x^3 - 4096) = 0 \][/tex]
[tex]\[ x = 0 \quad \text{or} \quad x^3 = 4096 \][/tex]
[tex]\[ x = 0 \quad \text{or} \quad x = \sqrt[3]{4096} = 16 \][/tex]
So, the points of intersection are [tex]\( (0, 0) \)[/tex] and [tex]\( (16, 64) \)[/tex].
### Step 2: Set Up and Evaluate the Integral
To find the area between the curves, we integrate the difference of the functions over the interval [tex]\( [0, 16] \)[/tex].
1. The upper function [tex]\( y = \sqrt{16x} \)[/tex] and the lower function [tex]\( y = \frac{x^2}{16} \)[/tex].
2. Set up the integral:
[tex]\[ \text{Area} = \int_{0}^{16} \left( \sqrt{16x} - \frac{x^2}{16} \right) \, dx \][/tex]
### Step 3: Solve the Integral
1. Simplify the integrand:
[tex]\[ \sqrt{16x} = 4\sqrt{x} \][/tex]
[tex]\[ \text{Area} = \int_{0}^{16} \left( 4\sqrt{x} - \frac{x^2}{16} \right) \, dx \][/tex]
2. Compute the definite integral:
[tex]\[ \int 4\sqrt{x} \, dx = \int 4x^{1/2} \, dx = 4 \cdot \frac{2}{3} x^{3/2} = \frac{8}{3} x^{3/2} \][/tex]
[tex]\[ \int \frac{x^2}{16} \, dx = \frac{1}{16} \int x^2 \, dx = \frac{1}{16} \cdot \frac{x^3}{3} = \frac{1}{48} x^3 \][/tex]
3. Evaluate the integrals from 0 to 16:
[tex]\[ \left[ \frac{8}{3} x^{3/2} \right]_{0}^{16} - \left[ \frac{1}{48} x^3 \right]_{0}^{16} \][/tex]
[tex]\[ \left( \frac{8}{3} (16)^{3/2} \right) - \left( \frac{1}{48} (16)^3 \right) \][/tex]
4. Calculate [tex]\( (16)^{3/2} \)[/tex] and [tex]\( (16)^3 \)[/tex]:
[tex]\[ (16)^{3/2} = \sqrt{16^3} = \sqrt{4096} = 64 \][/tex]
[tex]\[ (16)^3 = 4096 \][/tex]
5. Substitute back into the integrals:
[tex]\[ \frac{8}{3} \cdot 64 - \frac{1}{48} \cdot 4096 \][/tex]
[tex]\[ \frac{512}{3} - \frac{4096}{48} \][/tex]
[tex]\[ \frac{512}{3} - \frac{4096}{48} = \frac{512}{3} - \frac{4096}{48} \][/tex]
6. Simplify [tex]\(\frac{4096}{48}\)[/tex]:
[tex]\[ \frac{4096}{48} = \frac{1024}{12} = \frac{256}{3} \][/tex]
7. Subtract the fractions:
[tex]\[ \frac{512}{3} - \frac{256}{3} = \frac{256}{3} \][/tex]
Thus, the area of the region included between the parabolas is [tex]\(\boxed{\frac{256}{3}}\)[/tex] square units.
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