Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. y = x³, y = √ x about the x-axis V= ?

Answers

Answer 1

Answer:

Answer:

Step-by-step explanation:

To find the volume of the solid obtained by rotating the region bounded by the curves y=x3y=x3 and y=xy=x

about the x-axis, we'll use the method of cylindrical shells.

First, let's sketch the region bounded by these curves to better understand the shape. The intersection points of y=x3y=x3 and y=xy=x

are the points where x3=xx3=x

, which gives x=0x=0 and x=1x=1.

Now, we'll set up the integral to find the volume using cylindrical shells:

The volume VV can be calculated using the formula:

V=2π∫abx⋅(f(x)−g(x)) dxV=2π∫abx⋅(f(x)−g(x))dx

Where aa and bb are the bounds of integration (in this case, 00 and 11), and f(x)f(x) and g(x)g(x) are the heights of the shells. In this case, f(x)=x3f(x)=x3 and g(x)=xg(x)=x

So, the volume can be calculated as:

V=2π∫01x⋅(x3−x) dxV=2π∫01x⋅(x3−x

)dx

Now, simplify the integrand:

V=2π∫01(x4−xx) dxV=2π∫01(x4−xx

)dx

Split the integral into two parts:

V=2π∫01x4 dx−2π∫01xx dxV=2π∫01x4dx−2π∫01xx

Evaluate each integral separately:

V=2π[x55]01−2π[2x5/25/2]01V=2π[5x5]01−2π[5/22x5/2]01

V=2π(15)−2π(25)=2π5V=2π(51)−2π(52)=52π

So, the volume of the solid obtained by rotating the region bounded by y=x3y=x3 and y=xy=x

about the x-axis is 2π552π cubic units.

Answer 2

Answer:

The volume (V) of the solid obtained by rotating the region bounded by the curves $y = x^3\) and $y = √(x)$ about the x-axis is $\(V = (8)/(15)$ cubic units.$

To find the volume of the solid using the disk method, we integrate the cross-sectional area of each infinitesimally thin disk perpendicular to the x-axis.

The bounds of integration are determined by finding the x-values where the two curves intersect:

$x^3 = √(x) \implies x^6 = x \implies x^5 = 1 \implies x = 1.$

The radius of each disk is $r = x^3 - √(x)$ , and the area of each disk is $$A = \pi r^2 = \pi \left((x^3 - √(x))\right)^2$.$

The integral for the volume becomes:

$V = \int_(0)^(1) \pi \left((x^3 - √(x))\right)^2 \, dx.$

Evaluating this integral gives $$V = (8)/(15)$ cubic units.$

In summary, the volume of the solid obtained by rotating the region bounded by $y = x^3\) and $y = √(x)$ about the x-axis is $\(V = (8)/(15)$ cubic units.$ The volume is calculated by integrating the cross-sectional areas of infinitesimally thin disks formed by rotating the region.

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a Potter uses 3/5 of a pound of clay to make a bowl. how many bowls could the Potter make with 10 pounds of clay?

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Solve the triangle, find m∠A and m∠C. Round angles to the nearest degree.m∠A= __∘

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Answers

In the given right triangle ABC, m∠A ≈ 26.44° and m∠C ≈ 63.56°.

To solve the right triangle ABC, we can use trigonometric ratios. In a right triangle, the three main trigonometric ratios are:

1. Sine (sin): $\sin(\theta) = \frac{{\text{opposite side}}}{{\text{hypotenuse}}}$

2. Cosine (cos): $\cos(\theta) = \frac{{\text{adjacent side}}}{{\text{hypotenuse}}}$

3. Tangent (tan): $\tan(\theta) = \frac{{\text{opposite side}}}{{\text{adjacent side}}}$

Given:

AC = 38

AB = 17

To find the angles m∠A and m∠C, we can use the sine and cosine ratios, respectively.

1. For m∠A:

$\sin(m\angle A) = \frac{{AB}}{{AC}} = \frac{{17}}{{38}}\)\n\n\(m\angle A= \sin^(-1)\left(\frac{{17}}{{38}}\right)$

2. For m∠C:

$\cos(m\angle C) = \frac{{AB}}{{AC}} = \frac{{17}}{{38}}\)\n\n\(m\angle C = \cos^(-1)\left(\frac{{17}}{{38}}\right)$

Let's calculate the angles:

$1. $m\angle A \approx \sin^(-1)\left(\frac{{17}}{{38}}\right) \approx 26.44^\circ$\n\n2. $m\angle C \approx \cos^(-1)\left(\frac{{17}}{{38}}\right) \approx 63.56^\circ$$

Therefore, m∠A ≈ 26.44° and m∠C ≈ 63.56° (rounded to the nearest degree).

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Answer:

$m\angle A=63^\circ\nm\angle C=26^\circ$

Step-by-step explanation:

Trigonometric Ratios

The ratios of the sides of a right triangle are called trigonometric ratios. The longest side of the triangle is called the hypotenuse and the other two sides are called the legs.

Selecting any of the acute angles as a reference, it has an adjacent side and an opposite side. The trigonometric ratios are defined upon those sides.

The cosine ratio is defined as:

$\displaystyle \cos\theta=\frac{\text{adjacent leg}}{\text{hypotenuse}}$

Note the angle A of the figure has 17 as the adjacent leg and 38 as the hypotenuse, so we can directly apply the formula:

$\displaystyle \cos A=(17)/(38)$

$\cos A=0.4474$

Using a scientific calculator, we get the inverse cosine:

$A=\arccos(0.4474)$

$A\approx 63^\circ$

Since A+B+C=180°, we can solve for C:

C = 180° - A - B

C = 180° - 63° - 90°

C = 26°

Thus:

$m\angle A=63^\circ\nm\angle C=26^\circ$

A line crosses the y-axis at (0,4) and has a slope of -2. Find an equation for the this line.A) y = 2x + 4 B) y = -2x + 4 C) y = -2x - 4 D) y = -4x + 2 <-- 6)Write the equation of the line with the given slope and y-intercept.
slope = 5 y-intercept = 8 A) y = 5x + 8 B) x = 5y + 8 <-- C) y = 8x + 5 D) x = 8y + 5

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C) y = -2x -

A) y = 5x + 8

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