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2 years ago

Volumes of the solids with disk or shell method?

Mathematics

1 answer:

lina2011 [118]2 years ago

8 0

The answer to the questions of volumes are given as follows

a) $v=128 \pi$

b) $v=\frac{128}{3} \pi$

c) $v=\frac{1024}{5} \pi$

d) $V=\frac{13568}{15} \pi$

Generally, the questions are mathematically solved below

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

a) x-axis

if $y=4, x=16, x=0$

Using disk method

$} v=\pi \int_{0}^{16}(\sqrt{x})^{2} d x \\$

$v=\pi\left(\frac{x^{2}}{2}\right)_{0}^{16} \\$

$v=\frac{\pi}{2} \times 16 \times 16 \\$

$v=128 \pi$

b) line y=4

if x=0, y=0 ;

$y=\sqrt{x} \Rightarrow x=y^{2}$

Using shell method

$v = \int_{0}^{4} 2 \pi(4-y) \cdot y^{2} d y \\$

$v=2 \pi \int_{0}^{4}\left(4 y^{2}-y^{3}\right) d y$

$v=2 \pi\left[\frac{4 y^{3}}{3}-\frac{y^{4}}{4}\right]_{0}^{4} \\$

$v=\frac{2 \pi}{12}[1024-768] \\$

$v=\frac{512 \pi}{12} \\$

$v=\frac{128}{3} \pi$

c) y-axis

0 ≤ y ≤ 4

x=y^2

Using disk method

volume

$v=\pi \int_{0}^{4} y^{4} d y$$

$v=\pi\left(\frac{y 5}{5}\right)_{0}^{4} \\$

$v=\frac{1024}{5} \pi$

d) line x=-1

y=√x, y=4, x=0

0 ≤ x ≤ 6

Using shell method

volume is

$V=\int_{0}^{16} 2 \pi(1+x) \sqrt{x} d x$$

$V=2 \pi\int_{0}^{16}(x^{1 / 2}+x^{3 / 2}\right) )d x\right. \\$

$V=2 \pi\left[\frac{x^{3 / 2}}{3 / 2}+\frac{x^{5 / 2}}{5 / 2}\right]_{0}^{16} \\$

$V=2 \pi\left[2 / 3 \cdot\left(4^{2}\right)^{3 / 2}+2 / 5\left(4^{2}\right)^{5 / 2}\right] \\$

$V=4 \pi / 3 \cdot 4^{3}+4 \pi / 5 \cdot 4^{5} \\$

$V=4 \pi\left(\frac{1}{3}+\frac{4^{2}}{5}\right)$

$V=\frac{256}{15}(5+48) \pi \\$

$V=\frac{256 \times 53}{15} \pi \\$

$V=\frac{13568}{15} \pi$

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fiasKO [112]

Answer:

A is true.

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3 years ago

Find the radius of a circle using the Pythagorean theorem, given that the center is at (1, 5) and the point (5, 1) lies on the c

Assoli18 [71]

Given :

Center of circle, C = ( 1 , 5 )

Point on circle, P = ( 5 , 1 )

To Find :

The radius of a circle.

Solution :

We know, radius of circle is the distance between center and point of circle.

Distance between two points A(p,q) and B(r,s) is given by :

$D = \sqrt{(p-r)^2+(q-s)^2}$

So, distance between the points (1, 5) and (5, 1) is :

$D = \sqrt{(1-5)^2+(5-1)^2}\\\\D = 5.66 \ units$

Therefore, radius of circle is 5.66 units.

Hence, this is the required solution.

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3 years ago

Please help!!<br> Write a matrix representing the system of equations

frozen [14]

Answer:

$(4, -1, 3)$

Step-by-step explanation:

We have the system of equations:

$\left\{ \begin{array}{ll} x+2y+z =5 \\ 2x-y+2z=15\\3x+y-z=8 \end{array} \right.$

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

$\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}$

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

$\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}$

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

$\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}$

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

$\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}$

Now, we can multiply R₂ by -1/5. This yields:

$\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}$

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}$

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}$

We can now reduce R₃ by multiply it by -1/4:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

$\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

And finally, we can eliminate the second 1 by adding -(R₃):

$\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$