Question

Volume of a Bowl A bowl has a shape that can be generated by revolving the graph of \(y=x^{2} / 2\) between \(y=0\) and \(y=5\) about the \(y\) -axis. (a) Find the volume of the bowl. (b) If we fill the bowl with water at a constant rate of 3 cubic units per second, how fast will the water level in the bowl be rising when the water is 4 units deep?

Short answer

The volume of the bowl is \(25\pi\) cubic units and the rate at which the water level rises when the water is 4 units deep is \(3 / (8\pi)\) units per second.

Step-by-step solution

Find the volume of the bowl

The volume of the revolution is given by the integral formula \[V = \pi \int_{{a}}^{{b}} (r(y))^2 dy. \] The revolution is about the y-axis, so the radius at a given point in the y-axis would be \(r(y) = x = \sqrt{2y}\). So, the volume can be given as \[V = \pi \int_{{0}}^{{5}} ({\sqrt{2y}})^2 dy\]

Solve the integral

Solving the integral we get \[V = \pi \int_{{0}}^{{5}} 2y dy = \pi [y^2]_{{0}}^{{5}} = \pi (5^2 - 0^2) = 25\pi.\] So, the volume of the bowl is \(25\pi\) cubic units.

Set up the differential equation

The volume of water is rising at a rate of 3 cubic units per second which gives \(dV/dt = 3\). We are tasked to find \(dh/dt\) when \(h = 4\). Just like step 1, we know the volume of water filled is given by \[V = \pi \int_{{0}}^{{h}} 2y dy.\] Therefore, by differentiation, we get \[dV/dt = d(\pi [y^2]_{{0}}^{{h}})/dt = 2 \pi h (dh/dt).\]

Solve for the rate of height increase

Set up the equation for \(dh/dt\) when \(h = 4\) by substituting the given values into the differential equation \[3 = 2 \pi (4)(dh/dt)\] Solve the above equation to get \(dh/dt = 3 / (8\pi)\).

Key concepts

Integral Calculus

Integral calculus is a part of mathematics that involves finding the area under a curve. This process is crucial when determining the volume of objects, especially when they are not standard shapes. In the case of rotational volumes, integral calculus provides us with a way to calculate the exact volume by integrating over the curve that forms the shape.

For example, to find the volume of a bowl shaped like a parabola, we would use the formula \(V = \pi \int_{{a}}^{{b}} (r(y))^2 dy\), where \(r(y)\) is the distance from the point on the curve to the axis of rotation, in this case, the y-axis. By integrating from the base \(y=a\) to the top \(y=b\) of the bowl, we can calculate the total volume enclosed by the parabolic shape. This integrals application symbolizes the fundamental concept of accumulation - summing infinitesimally small products of the function value (the radius squared) and a small change in width along the y-axis (dy).

Solid of Revolution

A solid of revolution is a 3D object created by rotating a 2D shape around an axis. The resulting solid has a symmetrical form and is often observed in everyday objects, like bowls and vases. To calculate the volume of such objects, we employ the method of discs or washers, which involves slicing the solid into thin discs perpendicular to the axis of rotation and summing their volumes.

In the exercise, the bowl is formed by revolving the graph of \(y=x^2/2\) around the y-axis. Imagine slicing the bowl into small, horizontal layers or discs. Each disc has a tiny thickness (\(dy\)) and a radius that depends on \(y\), specifically \(\sqrt{2y}\). The volume of each disc is given by \(\pi r^2 dy\), and the total volume is the sum of all these individual discs' volumes — an integral from the bottom of the bowl, \(y=0\), to the top, \(y=5\).

Rate of Change

The concept of a 'rate of change' is a quantifiable measure that tells us how a certain quantity changes with respect to another. In our context, it relates to how fast a level changes over time, such as how quickly the water level rises in a bowl when it is filled at a constant rate. It's often denoted as \(\frac{dV}{dt}\) or \(\frac{dh}{dt}\) indicating the change in volume (\(V\)) or height (\(h\)) with time (\(t\)).

In the exercise's second part, we are given the rate at which the volume of water increases and are asked to find out how fast the water's height is rising when the water is 4 units deep. This applies differential calculus to relate the rate of volume increase to the rate of height increase by applying the mathematical relationship between volume and height of the water in the bowl.

Differential Calculus

Differential calculus is the study of how things change and is used to calculate the rate at which quantities change. This branch of calculus deals with derivatives, which represent rates of change. For instance, if we have an equation that describes a relationship between volume and height, the derivative tells us how quickly the volume changes as the height changes.

In our exercise, we used differential calculus to relate the volume of water in the bowl (as a function of time) to the height of the water. By differentiating the volume with respect to time, we obtained an equation that can tell us the rate at which the height of water is increasing when the bowl is being filled. After setting up the differential equation, we solved for \(\frac{dh}{dt}\) when \(h = 4\) to find the speed of the rising water level at that height, demonstrating the powerful application of differential calculus in solving real-world problems.