Surface Area and Integration Concepts

Surface Area and Integration Concepts

Assessment

Interactive Video

Mathematics

10th - 12th Grade

Hard

CCSS
8.G.C.9, 6.EE.C.9, HSG.GMD.A.3

Standards-aligned

Created by

Jackson Turner

FREE Resource

Standards-aligned

CCSS.8.G.C.9
,
CCSS.6.EE.C.9
,
CCSS.HSG.GMD.A.3
The video tutorial explains how to find the surface area of revolution about the y-axis for the function y = cube root of 6x over the interval x = 0 to 4/3. It discusses the graph of the function, verifies values, and explores integration methods with respect to x and y. The tutorial solves for x, finds derivatives, and performs integration using u-substitution to calculate the surface area. Finally, it provides the exact and approximate surface area values.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the function given for finding the surface area of revolution?

y = 6x cubed

y = square root of 6x

y = cube root of 6x

y = 6x squared

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the interval for x in the given problem?

[0, 3]

[0, 1]

[0, 4/3]

[0, 2]

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

When rotating a curve about the y-axis, what are the two possible variables for integration?

x and z

y and z

x and y

z and w

Tags

CCSS.6.EE.C.9

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main factor in deciding whether to integrate with respect to x or y?

The length of the interval

The complexity of f'(x) or f'(y)

The value of the function at endpoints

The graph of the function

Tags

CCSS.6.EE.C.9

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the expression for x in terms of y after solving the function y = cube root of 6x?

x = y^3 * 6

x = 6 / y^3

x = 6y^3

x = y^3 / 6

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the derivative f'(y) of the function f(y) = y^3 / 6?

y^2 / 2

y^3 / 2

3y^2

1/2 y^2

Tags

CCSS.8.G.C.9

CCSS.HSG.GMD.A.3

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What substitution is used to evaluate the integral for the surface area?

u = y^3 / 6

u = y^2 / 2

u = 1 + y^4 / 4

u = y^4 / 4

Tags

CCSS.8.G.C.9

CCSS.HSG.GMD.A.3

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