The region under the curves #x=0, x=y-y^4# is rotated about a) the x axis and b) the y axis. How do you sketch the region and find the volumes of the two solids of revolution?

Question

The region under the curves #x=0, x=y-y^4# is rotated about a) the x axis and b) the y axis.  How do you sketch the region and find the volumes of the two solids of revolution?

Charles Anctil · Accepted Answer

See the answer below:

Emilia Aguilar · Answer

undefined To sketch the region and find the volumes of the two solids of revolution:

a) For rotation about the x-axis, sketch the region bounded by the curves x = 0 and x = y - y^4. Then, integrate the formula for the volume of a solid of revolution with respect to x.

$$ V_x = \int_{a}^{b} \pi (f(x))^2 \, dx $$

In this case, $f(x) = y - y^4$. To find the limits of integration, set the equations equal to each other and solve for y.

$$ y - y^4 = 0 $$
$$ y(1 - y^3) = 0 $$
$$ y = 0 \text{ or } y = 1 $$

So, the limits of integration are from y = 0 to y = 1. The integral becomes:

$$ V_x = \pi \int_{0}^{1} (y - y^4)^2 \, dy $$

After integrating, you can find the volume of the solid of revolution about the x-axis.

b) For rotation about the y-axis, sketch the region bounded by the curves x = 0 and x = y - y^4. Then, integrate the formula for the volume of a solid of revolution with respect to y.

$$ V_y = \int_{c}^{d} \pi (f(y))^2 \, dy $$

In this case, $f(y) = y - y^4$. To find the limits of integration, determine the y-values at which the curves intersect.

$$ y - y^4 = 0 $$
$$ y(1 - y^3) = 0 $$
$$ y = 0 \text{ or } y = 1 $$

So, the limits of integration are from y = 0 to y = 1. The integral becomes:

$$ V_y = \pi \int_{0}^{1} (y - y^4)^2 \, dx $$

After integrating, you can find the volume of the solid of revolution about the y-axis.