As part of his work for NASA, Dr. Murdock was asked to find out what percentage of people in the continental united States saw H
1 answer:
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Answer:
a)
b)
c)
Explanation
In order to solve these problems, we must start by sketching a drawing of what the graph of the problem looks like, this will help us analyze the drawing better and take have a better understanding of the problem (see attached pictures).
a)
On part A we must build an integral for the volume of the torus by using the shell method. The shell method formula looks like this:
Where r is the radius of the shell, y is the height of the shell and dr is the width of the wall of the shell.
So in this case, r=x so dr=dx.
y is given by the equation of the circle of radius 3 centered at (5,0) which is:
when solving for y we get that:
we can now plug all these values into the shell method formula, so we get:
now there is a twist to this problem since that will be the formula for half a torus.Luckily for us the circle is symmetric about the x-axis, so we can just multiply this integral by 2 to get the whole volume of the torus, so the whole integral is:
we can take the constants out of the integral sign so we get the final answer to be:
b)
Now we need to build an integral equation of the torus by using the washer method. In this case the formula for the washer method looks like this:
where R is the outer radius of the washer and r is the inner radius of the washer and dy is the width of the washer.
In this case both R and r are given by the x-equation of the circle. We start with the equation of the circle:
when solving for x we get that:
the same thing happens here, the square root can either give you a positive or a negative value, so that will determine the difference between R and r, so we get that:
and
we can now plug these into the volume formula:
This can be simplified by expanding the perfect squares and when eliminating like terms we end up with:
c) We are going to solve the integral we got by using the washer method for it to be easier for us to solve, so let's take the integral:
This integral can be solved by using trigonometric substitution so first we set:
which means that:
from this, we also know that:
so we can set the new limits of integration to be:
and
so we can rewrite our integral:
which simplifies to:
we can further simplify this integral like this:
We can use trigonometric identities to simplify this so we get:
we can solve this by using u-substitution so we get:
and:
so when substituting we get that:
when integrating we get that:
when evaluating we get that:
which yields: