The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method

Question

x = 1 − y⁴    x = 0,     about the line x = 9

Answer 1

First observe that setting the two curves equal to each other gives us

1-y^{4} = 0 \Rightarrow -y^{4} = - 1

\Rightarrow y^{4} = 1

\Rightarrow y = \pm 1 (y must be a real number)

Next observe that

1-y^{4} > 0 \Rightarrow (1 - y^{2})(1 + y^{2}) > 0

\Rightarrow 1 - y^{2} > 0 (since 1 + y^{2} > 0)

\Rightarrow -y^{2} > -1

\Rightarrow y^{2} < 1

\Rightarrow |y| < 1

\Rightarrow -1 < y < 1 (i.e. in between our two points of intersection)

From these preceding observations, we can deduce that

inner radius = 9 - (1 - y^{4}) = 8 + y^{4}

outer radius = 9 - 0 = 9

cross-sectional area = A(y)

= \pi*(9)^{2} - \pi*(8 + y^{4})^{2}

= \pi*[9^{2} - (8 + y^{4})^{2}]

= \pi*[81 - (8^{2} + 2*8*y^{4} + (y^{4})^{2})]

= \pi*[81 - (64 + 16*y^{4} + y^{8})]

= \pi*[17 - 16*y^{4} - y^{8}]

Finally, we find that the volume of the resulting solid is

V = \int_{-1}^{1} A(y)~dy

= \int_{-1}^{1} \pi*[17 - 16*y^{4} - y^{8}]~dy

= \pi*\int_{-1}^{1}[17 - 16*y^{4} - y^{8}]~dy

= \pi*[17*y - 16*(\frac{1}{5}*y^{5}) - \frac{1}{9}*y^{9}]_{-1}^{1}

= \pi*[17*y - \frac{16}{5}*y^{5} - \frac{1}{9}*y^{9}]_{-1}^{1}

= \pi*[[17*1 - \frac{16}{5}*1^{5} - \frac{1}{9}*1^{9}]

- [17*(-1) - \frac{16}{5}*(-1)^{5} - \frac{1}{9}*(-1)^{9}]]

= \pi*[[17 - \frac{16}{5} - \frac{1}{9}]

- [-17 + \frac{16}{5} + \frac{1}{9}]]

= \pi*[[17 - \frac{16}{5} - \frac{1}{9}]

+ [17 - \frac{16}{5} - \frac{1}{9}]]

= \pi*[34 - \frac{32}{5} - \frac{2}{9}]

= \pi*[\frac{1530}{45} - \frac{288}{45} - \frac{10}{45}]

= \pi*[\frac{1232}{45}]

= \frac{1232*pi}{45}

Answer 2

x = 0 and x = 1 - y^4 meet when 1 - y^4 = 0 or 1 = y^4 or y = ±1.

In this problem, x is a function of y and we are rotating about x = 9, so we will use the disc method.

We will be subtracting the volume created by rotating 1 - y^4 about x = 9 from x = 0 about x = 9.

For x = 0, this is cylinder of height (1 - -1) =2 and radius 9, or volume πr²h = π9² * 2 = 162π

Then, using I[a, b] for the integral from a to b and E[a, b] for the evaluation from a to b

I[-1, 1] π(9 - (1 - y⁴))² dy = I[-1, 1] π(8 + y⁴))² dy = π I[-1,1] 64 + 16y⁴ + y⁸ dy =

π(64y + 16y⁵/5 + y⁹/9) E[-1,1] = 2π(64 + 16/5 + 1/9) = 134 28/45 π

Then, 162π - 134 28/45 π = 27 17/45 π

Answer 3

We use shell: V = 2π∫₀¹(9 - x)·2(1- x)^1/4dx ≈ 86.01

I use TI-84 to evaluate this integral.

Let we try using disk method: V = π∫_-1¹(9² - (9 - 1 + y⁴)²dy = π∫_-1¹ (81 - 64 - 16y⁴ - y⁸)dy =

π∫_-1¹(17 - 16y⁴ - y⁸)dy = 2π∫₀¹(17 - 16y⁴ - y⁸)dy = 2π(17y - 16y⁵/5 - y⁹/9)₀¹ = 2π(17 - 16/5 - 1/9) ≈ 86.01