Improper Integral

Category: Integral Calculus

"Published in Suisun City, California, USA"

Evaluate

Solution:

The given equation above is Improper Integral where the limits are - ∞ and + ∞. In this case, we can split the limits as follows

Consider

Since the lower limit is -∞, we can rewrite the above equation as follows

Consider

Since the upper limit is +∞, we can rewrite the above equation as follows

Therefore,

Since the given Improper Integral has a value, then the given Improper Integral is convergent. 

Word Problem - Wine Dilution

Category: Algebra

"Published in Newark, California, USA"

A Chinese wine merchant has a cask full of brandy. He draws 5 liters and fills the cask with water. Once again, he draws out another 5 liters and fills the cask with water. How many liters does the cask hold if there are now 16 liters of brandy left?

Solution:

The given word problem describes the dilution of a substance which is a brandy with water. A certain amount of brandy is drawn from the cask and replaced with the equal amount of water. Let's analyze the word problem as follows

Let x = be the amount of brandy in liters before it is drawn
      x = be the volume of the cask or container in liters

Amount of brandy at the first drawn

Amount of brandy at the second drawn

The terms inside the bracket is the amount of the brandy left after the first drawn.

Multiply both sides of the equation by x, we have

Using Quadratic Formula,

Choose the positive sign because the initial amount of brandy must be more than 16 liters

Therefore, the amount of brandy before it is drawn is 25 liters. This is also the volume or size of the cask. 

More Right Circular Cylinder Problem, 2

Category: Solid Geometry

"Published in Newark, California, USA"

A right cylindrical solid of altitude 6 in. has the cross section shown in the shaded portion of the figure. BEDG is a circle whose radius is OG. AFCG is a circle which is tangent to the larger circle at G. If AB = CD = 5 in. and EF = 9 in., find the volume of the cylinder.

Photo by Math Principles in Everyday Life

Solution:

The cross section of a right circular cylinder consists of two tangent circles with their common point at Point G. Let's further analyze and label the cross section of a right circular cylinder as follows

Photo by Math Principles in Everyday Life

We noticed that line segments AC and FG are the chords of a small circle that intersect at point O, which is also the center of a big circle. From Plane Geometry, we know that the product of two divided chords is equal to the product of other two divided chords. We can solve for the radius of a big circle as follows

The line segment OF is calculated as follows

The radius of a small circle is calculated as follows

The area of the shaded portion of the cross section of a right circular cylinder is calculated as follows

             Area of a base = Area of Big Circle - Area of Small Circle

Therefore, the volume of a right circular cylinder is

or