Study the scenario and complete the question(s) that follow(s):
Designing a Unique Aquarium Tank
As an engineer designing custom aquarium tanks, you are tasked with creating a uniquely shaped tank to fit a specific space. The tank’s profile is defined by the curves and , and it is rotated around the line to form a distinctive, bowl-like structure. To ensure the tank meets volume requirements for water capacity and aquatic life, you must calculate its total volume using integral methods.
Source: Mafemba, J. 2024.

Using the shell method, calculate the volume generated by rotating the region around the specified line in the scenario above.
(8 Marks)

Ask by Reeves Whittaker. in South Africa

Mar 28,2025

Question

Study the scenario and complete the question(s) that follow(s):
Designing a Unique Aquarium Tank
As an engineer designing custom aquarium tanks, you are tasked with creating a uniquely shaped tank to fit a specific space. The tank’s profile is defined by the curves and , and it is rotated around the line to form a distinctive, bowl-like structure. To ensure the tank meets volume requirements for water capacity and aquatic life, you must calculate its total volume using integral methods.
Source: Mafemba, J. 2024.

Using the shell method, calculate the volume generated by rotating the region around the specified line in the scenario above.
(8 Marks)

Ask by Reeves Whittaker. in South Africa

Mar 28,2025

UpStudy AI · Accepted Answer

**Step 1. Determine the Bounds and Express the Region**

The region is bounded by the curves  
$$
y = -\sqrt{x} + 2 \quad \text{and} \quad y = 1.
$$
To find their intersection, set:
$$
-\sqrt{x} + 2 = 1.
$$
Solve for $$ x $$:
$$
-\sqrt{x} = 1 - 2 = -1 \quad \Rightarrow \quad \sqrt{x} = 1 \quad \Rightarrow \quad x = 1.
$$
Thus, the region is described by $$ x $$ from 0 to 1 and, for each $$ x $$, $$ y $$ varies from 1 up to $$ y=-\sqrt{x}+2 $$.

It is more convenient for the shell method (when the axis of rotation is horizontal) to express the region in terms of $$ y $$. Solve $$ y=-\sqrt{x}+2 $$ for $$ x $$:
$$
y = -\sqrt{x} + 2 \quad \Rightarrow \quad \sqrt{x} = 2 - y \quad \Rightarrow \quad x = (2-y)^2.
$$
The $$ y $$-values in the region range from $$ y=1 $$ (the line) up to $$ y=2 $$ (when $$ x=0 $$).

For a fixed $$ y $$ between 1 and 2, the $$ x $$-coordinate runs from 0 to $$ (2-y)^2 $$.

**Step 2. Set Up the Shell Method Integral**

When we rotate the region around the line $$ y = -1 $$, we use horizontal shells. For a shell at a typical $$ y $$ (with $$ 1 \le y \le 2 $$):

- **Radius:** The distance from $$ y $$ to the axis $$ y = -1 $$ is  
  $$
  \text{radius} = y - (-1) = y + 1.
  $$
- **Height:** The horizontal length of the region at the height $$ y $$ is  
  $$
  \text{height} = (2-y)^2.
  $$
- **Thickness:** $$ dy $$.

Thus, the volume of a typical shell is given by
$$
dV = 2\pi \, (\text{radius}) \, (\text{height})\, dy = 2\pi (y+1)(2-y)^2 \, dy.
$$

**Step 3. Form the Integral**

The total volume is obtained by integrating from $$ y=1 $$ to $$ y=2 $$:
$$
V = 2\pi \int_{1}^{2} (y+1)(2-y)^2 \, dy.
$$

**Step 4. Simplify the Integrand**

First, expand $$(2-y)^2$$:
$$
(2-y)^2 = 4 - 4y + y^2.
$$
Then, multiply by $$ y+1 $$:
$$
(y+1)(4-4y+y^2) = y(4-4y+y^2) + 1(4-4y+y^2).
$$
Calculate each term:
- $$ y(4-4y+y^2) = 4y - 4y^2 + y^3 $$,
- $$ 4-4y+y^2 $$ remains the same.

Adding, we get:
$$
4y - 4y^2 + y^3 + 4 - 4y + y^2 = y^3 - 3y^2 + 4.
$$
Thus, the integrand simplifies to:
$$
(y+1)(2-y)^2 = y^3 - 3y^2 + 4.
$$

**Step 5. Evaluate the Integral**

The volume integral now becomes:
$$
V = 2\pi \int_{1}^{2} \left(y^3 - 3y^2 + 4\right) dy.
$$

Find the antiderivative:
$$
\int \left(y^3 - 3y^2 + 4\right) dy = \frac{y^4}{4} - y^3 + 4y.
$$

Evaluate at the bounds:

At $$ y = 2 $$:
$$
\frac{2^4}{4} - 2^3 + 4(2) = \frac{16}{4} - 8 + 8 = 4 - 8 + 8 = 4.
$$

At $$ y = 1 $$:
$$
\frac{1^4}{4} - 1^3 + 4(1) = \frac{1}{4} - 1 + 4 = \frac{1}{4} + 3 = \frac{13}{4}.
$$

Taking the difference:
$$
\left[\frac{y^4}{4} - y^3 + 4y\right]_{1}^{2} = 4 - \frac{13}{4} = \frac{16}{4} - \frac{13}{4} = \frac{3}{4}.
$$

Therefore, the volume is:
$$
V = 2\pi \cdot \frac{3}{4} = \frac{3\pi}{2}.
$$

**Final Answer**

$$
V = \frac{3\pi}{2}.
$$
The volume of the aquarium tank is $$ \frac{3\pi}{2} $$ cubic units.

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