Video Transcript
Use the slicing method to find the
volume of the solid whose base is the region inside the circle 𝑥 squared plus 𝑦
squared is equal to two centred at the origin, if the cross sections are equilateral
triangles perpendicular to the 𝑥-axis.
We’re asked to find the volume of a
solid whose base is the region inside the circle 𝑥 squared plus 𝑦 squared is equal
to two centred at the origin where the cross sections are equilateral triangles
perpendicular to the 𝑥-axis. Let’s first sketch our base and our
cross sections. Our base is the circle 𝑥 squared
plus 𝑦 squared is equal to two. This means the radius 𝑟 squared is
equal to two, which means that the radius 𝑟 is equal to plus or minus the square
root of two.
Our cross sections are equilateral
triangles perpendicular to the 𝑥-axis. The peak of our solid will be
traced out by the apexes of these triangles. Remember, we’ll want to calculate
the volume of the solid. And that’s equal to the definite
integral of the areas of the cross sections between limits defined by the boundaries
of the base. Our first task is then to calculate
the area of the triangles.
Remember that the area of a
triangle is half the base times the height. In our case, the base is two 𝑦
since it’s split in half by the 𝑥-axis. And since this is an equilateral
triangle, each side has length two 𝑦. Now, remember that because the
cross sections are perpendicular to the 𝑥-axis, we need to find the area as a
function of 𝑥. In order to do that, we’ll first
find the height ℎ in terms of 𝑦. And using the equation for our
base, we can find the area as a function of 𝑥.
To find the height ℎ, let’s use the
Pythagorean theorem. This says, for a right angle
triangle with sides 𝑎, 𝑏, 𝑐, 𝑎 squared plus 𝑏 squared is equal to 𝑐
squared. In our case, this means that ℎ
squared plus 𝑦 squared is equal to two 𝑦 all squared. That’s equal to four 𝑦
squared. So, ℎ squared is four 𝑦 squared
minus 𝑦 squared, which is three 𝑦 squared, which gives us that ℎ, since it’s a
length, is the positive square root of three times 𝑦.
Now, using our formula for the area
of a triangle, we have the area of our cross-sectional triangle is equal to half
times two 𝑦 times the square root of three times 𝑦. Cancelling our twos, that gives us
the area is equal to the square root of three times 𝑦 squared. Now, from the equation of our
circle, we have 𝑥 squared plus 𝑦 squared is equal to two. This means that 𝑦 squared is two
minus 𝑥 squared. So, if we substitute this into our
area, we get areas of function of 𝑥 so that our area is the square root of three
times two minus 𝑥 squared.
Remember that the volume is the
integral of the area where the limits are defined by the boundaries of our base. In our case, then, the limits are
negative root two to root two. So, our volume is the integral
between negative root two and root two of the square root of three times two minus
𝑥 squared with respect to 𝑥. As root three is a constant, we can
take this outside. So, our integral becomes root three
times the integral between negative root two and root two of two minus 𝑥 squared
with respect to 𝑥.
We can also use the symmetry of the
circle to make this a little bit easier. By the symmetry of the circle, we
can change the lower limit to zero and multiply the integral by two. So, we now have the volume equal to
two times the square root of three times the integral between zero and the square
root of two times two minus 𝑥 squared with respect to 𝑥. The integral of two with respect to
𝑥 is two 𝑥 evaluated between zero and square root of two. And the integral of 𝑥 squared with
respect to 𝑥 is one over three 𝑥 cubed evaluated between the limits zero and root
two.
Our volume is, therefore, two times
the square root of three times two 𝑥 minus 𝑥 cubed over three evaluated between
zero and root two. That gives us two times the square
root of three times two root two minus root two cubed over three minus zero minus
zero. Evaluating this, we get eight times
the square root of six over three.
The volume of the solid whose base
is the region inside the circle 𝑥 squared plus 𝑦 squared is two centred at the
origin where the cross sections are equilateral triangles perpendicular to the
𝑥-axis is eight times root six over three units cubed. We found this volume by finding the
areas of the cross-sectional triangles in terms of 𝑥. We then integrated this area over
an interval defined by the base.
