### Video Transcript

Set up an integral for the volume
of the solid obtained by rotating the region bounded by the curve π¦ is equal to π
raised to the power negative π₯ squared and the lines π¦ is zero, π₯ is equal to
negative five, and π₯ is equal to five about π¦ is equal to negative five.

In this question, weβre given a
region bounded by the curve π¦ is equal to π raised to the power negative π₯
squared, the line π¦ is equal to zero, and the two vertical lines π₯ is equal to
positive five and π₯ is equal to negative five. And we want to set up an integral
for the volume of the solid obtained by rotating this region about the line π¦ is
equal to negative five. This means that the line π¦ is
equal to negative five is our axis of rotation.

It can be helpful to try and sketch
the solid weβd like to find the volume of, but note that our two sketches have
different scales. Now, the solid formed by rotating
the given region has washers, that is, hollowed disks, as vertical cross
sections. The cross sections have both an
inner and an outer radius. Letβs call the outer radius π
sub
π and the inner radius π
sub πΌ. And now recalling that the area of
a circle with radius π is equal to ππ squared, then the area of our
cross-sectional disks π΄ sub π· is given by π times π
outer squared minus π times
π
inner squared. And taking π outside some
parentheses, thatβs π multiplied by π
outer squared minus π
inner squared.

Now, if we consider the outer
radius of one of our disks on our original diagram, the outer radius is the distance
from the center of rotation, thatβs the line π¦ is equal to negative five, and the
function π¦ is equal to π raised to the power negative π₯ squared, that is, π
raised to the power negative π₯ squared plus five. So our outer radius varies
depending on the value of π₯. And remember, π₯ ranges from
negative five to plus five. If we next consider the inner
radius π
πΌ, thatβs the distance between the line π¦ is equal to zero and π¦ is
equal to negative five, and thatβs five.

Now, to find the volume of this
solid, we take the infinite sum of the areas of all our cross-sectional disks
between π₯ is negative five and π₯ is plus five. That is, the volume is the integral
between negative five and five of the area of the disk with respect to π₯. And our integral is with respect to
π₯, since the axis of rotation is parallel to the π₯-axis. And so, for our volume, weβre
integrating π multiplied by the outer radius squared minus the inner radius squared
with respect to π₯ between negative five and five.

Weβre asked to set up an integral
for this volume. So all we need to do now is to work
out what the outer radius squared minus the inner radius squared is. And the outer radius squared is π
raised to the power negative π₯ squared plus five squared, that is, π raised to the
power negative two π₯ squared plus 10 multiplied by π raised to the power negative
π₯ squared plus 25. The inner radius squared is five
squared, and thatβs 25. And so we have the outer radius
squared minus the inner radius squared is π raised to the power negative two π₯
squared plus 10π raised to the power negative π₯ squared plus 25 minus 25. And since 25 minus 25 is zero, we
have the outer radius squared minus the inner radius squared is equal to π raised
to the power negative two π₯ squared plus 10π raised to the power negative π₯
squared.

Now, making some space, we can
substitute this into our integral. This gives us that our volume is
the integral between negative five and five of π multiplied by π raised to the
power negative two π₯ squared plus 10π raised to the power negative π₯ squared with
respect to π₯. And since we can take the factor of
π outside our integral, we have that the volume of the solid obtained by rotating
the region bounded by the curve π¦ is equal to π raised to the power negative π₯
squared and the lines π¦ is equal to zero, π₯ is negative five, and π₯ is positive
five about the line π¦ is equal to negative five is given by π multiplied by the
integral between π₯ is negative five and π₯ is positive five of π raised to the
power negative two π₯ squared plus 10 multiplied by π raised to the power negative
π₯ squared with respect to π₯.