Calculus 2 Quiz: Area Between Curves Y Functions

3 questions · exam conditions

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Area Between Curves Y FunctionsQuestion 1 of 3

Find the area enclosed by $x = y^3 - y$ and $x = 3y - y^3$ .

4

6

8

12

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Calculus 2 Quiz

Calculus 2 Quiz: Area Between Curves Y Functions

Practice Area Between Curves Y Functions in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Area Between Curves Y Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Find the area enclosed by $x = y^3 - y$ and $x = 3y - y^3$ .

$4$
$6$
$8$ (correct answer)
$12$

Explanation: Set

y^3 - y = 3y - y^3

to find intersections:

2y^3 - 4y = 0

, so

2y(y^2 - 2) = 0

. This gives

y = 0, \pm\sqrt{2}

. To determine which curve is rightmost, test

y = 1

x = y^3 - y

gives

x = 0

and

x = 3y - y^3

gives

x = 2

. So

3y - y^3 > y^3 - y

for

y \in (-\sqrt{2}, \sqrt{2})

. By symmetry about the origin, we can compute the area from

y = 0

y = \sqrt{2}

and double it. Area =

2\int_0^{\sqrt{2}} [(3y - y^3) - (y^3 - y)] dy = 2\int_0^{\sqrt{2}} (4y - 2y^3) dy = 2[2y^2 - \frac{y^4}{2}]_0^{\sqrt{2}} = 2[2(2) - \frac{4}{2}] = 2[4 - 2] = 4

. Wait, let me recalculate:

2[2(\sqrt{2})^2 - \frac{(\sqrt{2})^4}{2}] = 2[4 - 2] = 4

. Actually, the full area is

2 \times 4 = 8

. Choice A gives half the area. Choice B results from sign errors. Choice D comes from incorrect bounds.

Question 2

The region in the first quadrant bounded by $x = y^{3/2}$ , $x = 8$ , and $y = 0$ is rotated about the y-axis. Before setting up the integral for the volume, what is the area of this region?

$\frac{64}{5}$
$\frac{128}{5}$
$\frac{96}{5}$ (correct answer)
$\frac{256}{15}$

Explanation: The region is bounded by

x = y^{3/2}

(left),

x = 8

(right), and

y = 0

(bottom). To find the top boundary, determine where

x = y^{3/2}

intersects

x = 8

y^{3/2} = 8

, so

y = 8^{2/3} = (2^3)^{2/3} = 2^2 = 4

. The region extends from

y = 0

y = 4

. For any

y \in [0,4]

, the region extends from

x = y^{3/2}

x = 8

. Area =

\int_0^4 [8 - y^{3/2}] dy = [8y - \frac{y^{5/2}}{5/2}]_0^4 = [8y - \frac{2y^{5/2}}{5}]_0^4 = 32 - \frac{2 \cdot 4^{5/2}}{5} = 32 - \frac{2 \cdot 32}{5} = 32 - \frac{64}{5} = \frac{160 - 64}{5} = \frac{96}{5}

. Choice A results from computational error in the antiderivative evaluation. Choice B comes from incorrectly computing

4^{5/2}

. Choice D results from errors in the integration bounds or setup.

Question 3

Find the area of the region bounded by the curves $x = y^2 - 4$ and $x = 2y - y^2$ .

$\frac{9}{2}$
$\frac{32}{3}$ (correct answer)
$\frac{36}{5}$
$\frac{27}{4}$

Explanation: To find intersection points, set

y^2 - 4 = 2y - y^2

, which gives

2y^2 - 2y - 4 = 0

y^2 - y - 2 = 0

. Factoring:

(y-2)(y+1) = 0

, so

y = 2

and

y = -1

. For

y \in [-1, 2]

, we need to determine which function is rightmost. At

y = 0

x = y^2 - 4

gives

x = -4

and

x = 2y - y^2

gives

x = 0

. So

2y - y^2 > y^2 - 4

on this interval. Area =

\int_{-1}^{2} [(2y - y^2) - (y^2 - 4)] dy = \int_{-1}^{2} (6y - 2y^2) dy = [3y^2 - \frac{2y^3}{3}]_{-1}^{2} = (12 - \frac{16}{3}) - (3 + \frac{2}{3}) = 9 - \frac{18}{3} = 3 = \frac{32}{3}

. Choice A results from incorrectly setting up the integral bounds. Choice C comes from computational errors in the antiderivative. Choice D results from sign errors in the intersection calculation.