Question

find the volume of the solid obtained by rotating the region enclosed by: x=0 , y=1, x= y^4, about the line y=1 using the method of disks or washers.

Answer 1

Answer:

$V=\dfrac{1}{15}\pi$

Step-by-step explanation:

To find the volume of the solid obtained by rotating the region enclosed by the curves x = 0, y = 1, x = y⁴ about the line y = 1, we can use the method of washers.

The washer method formula for volume (V) is given by:

$\displaystyle V = \pi \int_a^b \left[ (f(x))^2 - (g(x))^2 \right] \;dx$

where:

• f(x) represents the outer radius.
• g(x) represents the inner radius.
• a and b are the limits of integration along the x-axis.

The limits of integration are a = 0 and b = 1, since the three lines intersect at (0, 0), and y = 1 and x = y⁴ intersect at (1, 1).

The outer radius f(x) is the distance from the axis of rotation to the outer curve, and the inner radius g(x) is the distance from the axis of rotation to the inner curve.  

In this case, the outer curve is y = 1 and the inner curve is x = y⁴ on the interval 0 ≤ x ≤ 1. So, as we are rotating about y = 1, then:

$f(x) = 1-1=0$

$g(x) = 1-x^{\frac14}$

Note that we have rewritten x = y⁴ as a function of x. We take the positive  quartic root, since we are rotating the region enclosed by the curve and y = 1.

Therefore, the integral for the volume using the washer method is:

$\displaystyle V = \pi \int_0^{1} \left[ (0)^2 - \left(1-x^{\frac14}\right)^2 \right] \;dx\\\\\\V = \pi \int_0^{1} \left(1-x^{\frac14}\right)^2 \;dx$

Evaluate the integral:

$\displaystyle V = \pi \int_0^{1} \left(1-2x^{\frac14}+x^{\frac12}\right) \;dx$

$V = \pi \left[x-\dfrac{2x^{\frac14+1}}{\frac14+1}+\dfrac{x^{\frac12+1}}{\frac12+1}\right]^1_0$

$V = \pi \left[x-\dfrac{2x^{\frac54}}{\frac54}+\dfrac{x^{\frac32}}{\frac32}\right]^1_0$

$V = \pi \left[x-\dfrac{8x^{\frac54}}{5}+\dfrac{2x^{\frac32}}{3}\right]^1_0$

$V=\pi \left[\left((1)-\dfrac{8(1)^{\frac54}}{5}+\dfrac{2(1)^{\frac32}}{3}\right)-\left((0)-\dfrac{8(0)^{\frac54}}{5}+\dfrac{2(0)^{\frac32}}{3}\right)\right]$

$V=\pi \left[\left(1-\dfrac{8}{5}+\dfrac{2}{3}\right)-\left(0-0+0\right)\right]$

$V=\dfrac{1}{15}\pi$