Help needed with Calculus

Question

Which of the following integrals will find the volume of the solid that is formed when the region bounded by the graphs of y = e^x, x = 2, and y = 1 is revolved around the line y = -1.

Options:

1)
2)
3)
4)

Answer 1

Hello! With solids of revolution, it's important to have a clear understanding of what the region in question looks like, and how it's being rotated.

In this case, if you graph y = e^x and y = 1, you will see that they intersect at the point (0,1). This forms the lefthand boundary of the region. The region is also bounded on the right side by the line x=2. The resulting region is somewhat triangular and is entirely above the x-axis.

Now, what does the solid look like if the region is rotated about the line y = -1? Well if you imagine rotating just the graph of y=e^x about the line y=-1, it would look like a sort of funnel, which is completely filled in. However, we are also rotating the graph of y=1. This creates a sort of hollowed out funnel, with a hole drilled through the middle.

Integration is all about adding up infinitely many pieces to form a whole. In this case, we can add up infinitely many tiny volumes to form the volume of the whole solid. Imagine slicing the solid vertically into smaller and smaller pieces; what does each piece start to look like? As the pieces get thinner and thinner, each one starts to look like a flat washer. This means that we can approximate the volume of an infinitely small piece of the volume as the surface area of a washer: the area of the larger circle minus the area of the hole (a smaller circle).

Now what we need to do is find the area of each "washer" at a given point x. In order to do that, we need to find an expression for the larger circle, and an expression for the smaller circle (the "hole"). We know that the area of a circle is given by π(radius)². In this case, the radius R of the larger circle is the distance between the outer function and the axis of rotation. The radius r of the smaller circle is the distance between the inner function and the axis of rotation. So:

R = e^x - (-1)

= e^x + 1

and

r = 1 - (-1)

= 2

Therefore, the area of each washer/cross-section is:

A = πR² - πr²

= π(e^x + 1)² - π(2)²

= π ((e^x + 1)² - (2)²)

Finally, we integrate across all values of x in the range [0,2] to find the total volume of the solid:

V = π ∫ (e^x + 1)² - (2)² dx

I hope this was helpful. Feel free to send any follow up questions!