Question

Work Let \(R\) be the region in the first quadrant bounded by the curve \(y=\sqrt{x^{4}-4},\) and the lines \(y=0\) and \(y=2 .\) Suppose a tank that is full of water has the shape of a solid of revolution obtained by revolving region \(R\) about the \(y\) -axis. How much work is required to pump all the water to the top of the tank? Assume \(x\) and \(y\) are in meters.

Short answer

Based on the given exercise to calculate the amount of work required to pump all the water out of a tank, we went through the following steps: 1. Find the intersection points of the curve and the bounded lines to determine the integration bounds. 2. Calculate the volume of the tank by using the disk method formula. 3. Calculate the work required to pump the water out of the tank by integrating the force exerted on each slice of water multiplied by the distance it is needed to be moved. After following these steps, we found that the work required to pump all the water to the top of the tank is approximately 112,023.7 Joules.

Step-by-step solution

Find the intersection points of the curve and thebounded lines.

To find the upper and lower bounds for our integrals, we need to find the intersection points of the bounding curve and lines. To do this, set \(y = \sqrt{x^4 - 4}\) equal to 0 and 2, and solve for the corresponding values of \(x\). When \(y=0\): \(0 = \sqrt{x^4 - 4} \Rightarrow {x^4 - 4 = 0} \Rightarrow x^4 = 4 \Rightarrow x = \sqrt[4]{4}\) When \(y=2\): \(2 = \sqrt{x^4 - 4} \Rightarrow x^4 = 4 + 2^2 = 8 \Rightarrow x = \sqrt[4]{8}\) The integration bounds will now be \(x=\sqrt[4]{4}\) and \(x=\sqrt[4]{8}\).

Calculate the volume of the tank.

The volume of the tank can be found using the disk method formula: \(V = \pi \int_a^b [f(x)]^2 dx\) where \(f(x) = y = \sqrt{x^4 - 4}\) and \(a=\sqrt[4]{4}, b=\sqrt[4]{8}\) We have: \(V = \pi \int_{\sqrt[4]{4}}^{\sqrt[4]{8}} (x^4 - 4) dx\) Now, integrate: \(V = \pi \left [ \frac{x^5}{5} - 4x \right ]_{\sqrt[4]{4}}^{\sqrt[4]{8}}\) Plug in the bounds to find the volume: \(V = \pi \left [ \left(\frac{(\sqrt[4]{8})^5}{5} - 4(\sqrt[4]{8})\right) - \left(\frac{(\sqrt[4]{4})^5}{5} - 4(\sqrt[4]{4})\right) \right ]\) \(V \approx 70.078\text{ m}^3\)

Calculate the work required to pump the water out of the tank.

The work required to pump the water out of the tank can be calculated by integrating the force exerted on each slice of water multiplied by the distance it is needed to be moved. The force to move a slice of water is its weight, which can be found as the product of its volume and the density of water (1000 kg/m^3). The distance each slice needs to travel is equal to its distance from the top of the tank 2 - y. The differential work equation is: \(dW =[1000\pi (x^4 - 4) ](2 - y) dy\) Since \(y = \sqrt{x^4 - 4}\), we can substitute to solve in terms of \(y\). \(dW = [1000\pi (y^2 + 4) ](2 - y) dy\) Now, we need to change the bounds of integration from \(x\) to \(y\). When \(x=\sqrt[4]{4}\), \(y = 0\). When \(x=\sqrt[4]{8}\), \(y = 2\). The total work can be found by integrating the differential work equation: \(W = \int_0^2 [1000\pi (y^2 + 4) ](2 - y) dy\) \(W = 1000\pi\int_0^2 (2y^2 + 8 - y^3 - 4y) dy\) Now, integrate and evaluate: \(W = 1000\pi [ \frac{2y^3}{3} + 8y - \frac{y^4}{4} - 2y^2 ]_0^2\) \(W \approx 112023.7 \text{ J}\) So, the work required to pump all the water to the top of the tank is approximately 112,023.7 Joules.

Key concepts

Solid of Revolution

A solid of revolution is formed by rotating a two-dimensional shape around an axis. In this problem, we're dealing with the region given by the curve \(y = \sqrt{x^4 - 4}\), bounded by the lines \(y = 0\) and \(y = 2\). When this region is revolved around the \(y\)-axis, it creates a three-dimensional shape, known as a solid of revolution.
This concept is essential in calculus because it helps us calculate properties like volume and surface area of complex shapes in a straightforward manner. By understanding how to revolve a shape around an axis, you can easily transition from dealing with two-dimensional calculus problems to solving three-dimensional ones.
This technique is particularly useful in engineering and physics, where rotational symmetry can simplify problems involving solid materials, like tanks or machine parts.

Integration by Parts

Integration by parts is a powerful technique in calculus used to solve integrals where the standard methods (like substitution) are not applicable. Although primarily used for products of functions, understanding integration by parts is crucial when dealing with more complex integrals like those found in problems involving the volume of revolution.
The formula for integration by parts is derived from the product rule of differentiation and is expressed as:

• \(\int u \, dv = uv - \int v \, du\)

Here, \(u\) and \(dv\) are parts of the integrand that you choose based on simplifying the resulting integral.
In the context of calculating work or volume of solids, integration by parts allows you to separate difficult integrals into parts that are easier to evaluate. Even if it's not directly applied in the specific solution steps above, having a strong grasp of this method can significantly ease the process of working through other calculus problems.

Volume of Revolution

The volume of a solid of revolution can be calculated using integration. In our problem, we used the disk method, which involves integrating the area of infinitely thin disks stacked along the axis of revolution. This method is particularly useful when revolving around one of the coordinate axes.
For a function \(f(x)\) revolved around the y-axis, the volume \(V\) can be expressed as:

• \(V = \pi \int_a^b [f(x)]^2 \ dx\)

Here, \(a\) and \(b\) are the bounds of the region in the x-direction.
It's also essential to convert all terms into the variable of integration. This ensures that the entire volume is calculated correctly. In this problem, switching from \(x\) to \(y\) as needed ensures clarity and accuracy in calculations.
Understanding how to apply this formula allows you to tackle a wide range of problems involving solids of revolution, from calculating the volume of vases and containers to understanding natural phenomena in a more mathematical light.