Calculate the area of the region bounded by y=6x and y=6x^2. And find the volume of the solid generated by revolving the region about the x-axis

Answers

Answer 1

Answer:

Hence, the Area is [tex]\frac{24}{5} \;\text {and the Volume of the bounded region is } \frac{24\pi}{5}[/tex]

Step-by-step explanation:

Concept :

Area of the region bounded by two curve :

[tex]\int_a^b(R^2-r^2)dx[/tex]

volume of the solid generated by revolving the region about the x-axis is :

 [tex]\int_a^b\pi(R^2-r^2)dx[/tex]   where [tex]R[/tex] is long radius of bounded region from x-axis and [tex]r[/tex] is short radius of bounded region from x-axis.

Given :

[tex]y=6x[/tex] and [tex]y=6x^2[/tex]

To find :

Area of the region bounded by [tex]y=6x[/tex] and [tex]y=6x^2[/tex] and volume of the solid generated by revolving the region about the x-axis.

Explanation :

[tex]\because y=6x\;\text{and}\;y=6x^2[/tex]

[tex]\therefore 6x=6x^2\\\Rightarrow x=1[/tex]

Area of the region bounded by [tex]y=6x[/tex] and [tex]y=6x^2[/tex] is :

     [tex]\int_a^b(R^2-r^2)dx=\int_0^1((6x)^2-(6x^2)^2)dx\\[/tex]

[tex]\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=\int_0^1(36x^2-36x^4)dx\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\int_0^1(x^2-x^4)dx\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\left(\frac{x^3}{3}-\frac{x^5}{5}\right)_0^1\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\left(\frac{1}{3}-\frac{1}{5}\right)\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=36\times\frac{2}{15}\\\Rightarrow \int_0^1((6x)^2-(6x^2)^2)dx=\frac{24}{5}[/tex]

Volume of the solid generated by revolving the region about the x-axis.

      [tex]\int_0^1\pi((6x)^2-(6x^2)^2)dx=\int_0^1\pi(36x^2-36x^4)dx[/tex]

[tex]\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\int_0^1(x^2-x^4)dx\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\left(\frac{x^3}{3}-\frac{x^5}{5}\right)_0^1\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\left(\frac{1}{3}-\frac{1}{5}\right)\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=36\pi\times\frac{2}{15}\\\Rightarrow \int_0^1\pi((6x)^2-(6x^2)^2)dx=\frac{24\pi}{5}[/tex]  


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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

[tex]t = 0.375s[/tex]

Step-by-step explanation:

Given

[tex]h(t) = -16t^2 + 4t + 33[/tex] --- driver 1

[tex]Rate = 2ft/s[/tex] -- driver 2

[tex]height = 33ft[/tex]

Required

The time they passed each other

First, we determine the function of driver 2.

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[tex]Rate = 2ft/s[/tex] and [tex]height = 33ft[/tex]

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[tex]h_2(t) = Height - Rate * t[/tex]

[tex]h_2(t) = 33 - 2t[/tex]

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Collect like terms

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Answer:

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Answer:

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A dime is dropped from a building at a
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

The image of the question can be seen below.

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Answers

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Step-by-step explanation:

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