Graph tis on a graph<br><br> y+1=1/3(x−3)

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago

How can you find the area of a circle if you only know the circumference?

AnnyKZ [126]

Https://youtu.be/3y1U0sntOsI

7 0

3 years ago

John got 24 out of 25 total questions correct on his test. What percent did he score on his test?

mamaluj [8]

Answer:

24 out of 25 is equal to 96%

96% = A

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

1. For a 30 student speech course, in how many ways can the student be selected to give the first 5 speeches?

liraira [26]

Combination problem:

1) 30 C 5 = 142506

2) 13 C 4 = 715

8 0

3 years ago

A ball thrown in the air has a height of y = - 16x² + 50x + 3 feet after x seconds. a) What are the units of measurement for the

Lubov Fominskaja [6]

<h2>Answer:</h2>

(a) ft/s

(b) 1ft/s

<h2>Step-by-step explanation:</h2>

Given equation;

y = (- 16x² + 50x + 3)ft -------------(i)

Where;

y is measured in feet(ft)

x is measured in seconds(s).

(a) The rate of change of y with respect to x is found by dividing the total change in y by the total change in x. i.e

Δy / Δx

Where;

Δy = y₂ - y₁

Δx = x₂ - x₁

∴ Δy / Δx = $\frac{y_2 - y_1}{x_2 - x_1}$ --------------(ii)

Since y is measured in feet, Δy will also be measured in feet.

Also, since x is measured in seconds, Δx will also be measured in seconds.

Therefore, the rate of change of y with respect to x (Δy / Δx) will be measured in feet per second (ft/s)

(b) The rate of change of y between x = 0 and x = 2 can be found by using equation (ii)

Where;

y₂ is the value of y at x = 2 found by substituting x = 2 into equation (i)

y₁ is the value of y at x = 0 found by substituting x = 0 into equation (i)

=> y₂ = - 16(2)² + 50(2) + 3 = 39

=> y₁ = - 16(1)² + 50(1) + 3 = 37

Now, substitute the values of y₂, y₁, x₂ and x₁ into equation (ii)

Δy / Δx = $\frac{39 - 37}{2 - 0}$

Δy / Δx = $\frac{2}{2}$

Δy / Δx = 1

Therefore, the rate of change of y is 1 ft/s

3 0

2 years ago

Graph tis on a graph y+1=1/3(x−3)