'Find the volume of the solid generated by revolving the region bounded by x=16-y^2, x-axis, and y-axis around the x-axis.'
I know in theory how to do it, but I'm a little confused by what constitutes the bounded region because it's a sideways parabola, but if the x-axis is a bound, does that mean only half of it counts?
So far I have that Volume equals pi times the integral from 0 to 16, but then getting the integrand conuses me.
Thamks!
confuses* my apologies
If you are using a limit of 16, then you must be integrating on x, so using discs of thickness dx,
v = ∫[0,16] πr^2 dx
where r = y = √(16-x)
v = ∫[0,16] π(16-x) dx = 128π
Or, you could use shells of thickness dy, where
v = ∫[0,4] 2πrh dy
where r=y and h=x = 16-y^2
v = ∫[0,4] 2πy(16-y^2) dy = 128π
To find the volume of the solid generated by revolving the region bounded by the curve \(x = 16 - y^2\), the x-axis, and the y-axis around the x-axis, you need to set up the integral correctly.
First, let's understand the bounded region. The equation \(x = 16 - y^2\) represents a sideways parabola that opens to the left. When y = 0, the x-coordinate is 16. As y increases or decreases, the x-coordinate decreases. So, the region bounded by \(x = 16 - y^2\), the x-axis, and the y-axis lies to the left of the y-axis and extends to the left indefinitely.
Since we are revolving this region around the x-axis, we need to consider a cross-section parallel to the x-axis. This cross-section will be a disk. As we move from left to right along the x-axis, the radius of the disk changes, so the volume can be calculated by integrating the areas of these disks.
To set up the integral, we need to express the radius as a function of x. To do this, we solve the equation \(x = 16 - y^2\) for y:
\[y^2 = 16 - x \]
\[y = \sqrt{16 - x} \]
The radius (r) of the disk at any x-coordinate is given by the distance between the x-axis and the curve \(y = \sqrt{16 - x}\). So, the radius is \(r = \sqrt{16 - x}\).
The height (h) of the disk can be considered as an infinitesimally small change in x, denoted as dx. Since the region is between the x-axis and the curve, the height corresponds to dx.
The volume of the disk is given by \(dV = \pi r^2 dx\).
To find the total volume, we integrate the volume of each disk over the interval where the bounded region lies. In this case, the bounded region lies between x = 0 to x = 16.
So, the integral for calculating the volume becomes:
\[V = \int_{0}^{16} \pi (\sqrt{16 - x})^2 dx \]
Simplifying, we have:
\[V = \int_{0}^{16} \pi (16 - x) dx \]
Evaluating this integral will give you the volume of the solid generated by revolving the region bounded by \(x = 16 - y^2\), the x-axis, and the y-axis around the x-axis.