Find the volume of the solid
formed by revolving the region
bound by the graphs of
, and about the -axis.
Use the disk method.

Round your answer to the nearest thousandth.

Ask by Hamilton Hardy. in the United States

Mar 23,2025

Question

Find the volume of the solid
formed by revolving the region
bound by the graphs of
, and about the -axis.
Use the disk method.

Round your answer to the nearest thousandth.

Ask by Hamilton Hardy. in the United States

Mar 23,2025

UpStudy AI · Accepted Answer

We first express the curve in a form that makes it easy to use the disk method. The region is bounded by

- $$y = \sqrt{4-x}$$
- $$x = 0$$
- $$y = 0$$

Since the curve is given as $$y = \sqrt{4-x}$$, we solve for $$x$$ to express the boundary in terms of $$y$$:

$$
y = \sqrt{4-x} \quad \Longrightarrow \quad x = 4-y^2.
$$

When we revolve the region about the $$y$$-axis, for a fixed $$y$$ the horizontal slice runs from $$x=0$$ to $$x=4-y^2$$. This slice, when revolved about the $$y$$-axis, forms a disk (since the inner radius is zero) with

$$
\text{Radius} = 4-y^2.
$$

The area of one such disk is

$$
A(y) = \pi \bigl(4-y^2\bigr)^2.
$$

The $$y$$-values range from the bottom of the region $$y=0$$ up to the top where the curve meets the $$y$$-axis. When $$x=0$$, the curve gives

$$
y = \sqrt{4-0} = 2.
$$

Thus, the limits for $$y$$ are $$y=0$$ and $$y=2$$.

The volume of the solid is then given by

$$
V = \int_{0}^{2} A(y) \, dy = \pi \int_{0}^{2} \left(4-y^2\right)^2 dy.
$$

We next expand the integrand:

$$
(4-y^2)^2 = 16 - 8y^2 + y^4.
$$

So the volume integral becomes

$$
V = \pi \int_{0}^{2} \left(16 - 8y^2 + y^4\right) dy.
$$

Now, we integrate term by term:

$$
\int_{0}^{2} 16 \, dy = 16y \Big|_0^2 = 16(2) - 16(0) = 32,
$$

$$
\int_{0}^{2} 8y^2 \, dy = 8\left(\frac{y^3}{3}\right) \Big|_0^2 = \frac{8(2^3)}{3} = \frac{64}{3},
$$

$$
\int_{0}^{2} y^4 \, dy = \frac{y^5}{5} \Big|_0^2 = \frac{2^5}{5} = \frac{32}{5}.
$$

Putting everything together:

$$
V = \pi \left[ 32 - \frac{64}{3} + \frac{32}{5} \right].
$$

To combine these terms, we use a common denominator, which is $$15$$:

$$
32 = \frac{480}{15}, \qquad \frac{64}{3} = \frac{320}{15}, \qquad \frac{32}{5} = \frac{96}{15}.
$$

Thus,

$$
32 - \frac{64}{3} + \frac{32}{5} = \frac{480}{15} - \frac{320}{15} + \frac{96}{15} = \frac{480 - 320 + 96}{15} = \frac{256}{15}.
$$

So, the volume is

$$
V = \pi \cdot \frac{256}{15} = \frac{256\pi}{15}.
$$

Now, rounding to the nearest thousandth, we approximate:

$$
V \approx \frac{256 \times 3.141593}{15} \approx \frac{804.248}{15} \approx 53.617.
$$

Thus, the volume of the solid is

$$
V = \frac{256\pi}{15} \quad \text{or approximately} \quad 53.617.
$$
The volume of the solid is $$ \frac{256\pi}{15} $$ or approximately 53.617 cubic units.

Find the volume of the solid
formed by revolving the region
bound by the graphs of
, and about the -axis.
Use the disk method.

Round your answer to the nearest thousandth.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Calculus Questions

Find the volume of the solid formed by revolving the region bound by the graphs of , and about the -axis. Use the disk method. Round your answer to the nearest thousandth.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Beyond the Answer

preguntas relacionadas

Latest Calculus Questions

Find the volume of the solid
formed by revolving the region
bound by the graphs of
, and about the -axis.
Use the disk method.

Round your answer to the nearest thousandth.