Math / Calculus

QuestionQ9. Find the volume of the solid object having base in the $x y$ -plane bounded by the curves $y=x^{2}-1$ and $y=1-|x|$ (see image below) and having cross-sections perpendicular to the $x$ -axis that are squares.

Studdy Solution

STEP 1

What is this asking? Find the volume of a 3D shape with a base between two curves, where slices perpendicular to the $x$ -axis are squares. Watch out! Don't mix up the curves; make sure you know which is on top and which is on the bottom!

STEP 2

1. Find the intersection points
2. Determine the side length of the square
3. Set up the integral for volume
4. Calculate the volume

STEP 3

First, we need to find where the curves intersect.
The curves are $y = x^2 - 1$ and $y = 1 - |x|$ .
Set them equal to each other to find the $x$ -values where they meet.

STEP 4

Let's solve $x^2 - 1 = 1 - |x|$ .
We need to consider two cases for $|x|$ : when $x \geq 0$ and when $x < 0$ .

STEP 5

For $x \geq 0$ , $|x| = x$ .
So, we solve: \[ x^2 - 1 = 1 - x $ Rearrange to: \[ x^2 + x - 2 = 0 $ Factor it: \[ (x + 2)(x - 1) = 0 $ So, $x = -2$ or $x = 1$.
Since $x \geq 0$ , we take $x = 1$ .

STEP 6

For $x < 0$ , $|x| = -x$ .
So, we solve: \[ x^2 - 1 = 1 + x $ Rearrange to: \[ x^2 - x - 2 = 0 $ Factor it: \[ (x - 2)(x + 1) = 0 $ So, $x = 2$ or $x = -1$.
Since $x < 0$ , we take $x = -1$ .

STEP 7

The intersection points are $x = -1$ and $x = 1$ .

STEP 8

The side length of the square is the distance between the curves at any $x$ .
The top curve is $y = 1 - |x|$ and the bottom curve is $y = x^2 - 1$ .

STEP 9

The side length $s$ is: \[ s = (1 - |x|) - (x^2 - 1) $ Simplify it: \[ s = 2 - |x| - x^2 $

STEP 10

The volume $V$ of the solid is the integral of the area of the square cross-sections from $x = -1$ to $x = 1$ .

STEP 11

The area of each square is $s^2$ .
So, the volume is: \[ V = \int_{-1}^{1} (2 - |x| - x^2)^2 \, dx $

STEP 12

Let's break the integral into two parts because of the absolute value: from $-1$ to $0$ and from $0$ to $1$ .

STEP 13

For $x < 0$ , $|x| = -x$ .
The integral becomes: \[ \int_{-1}^{0} (2 + x - x^2)^2 \, dx $

STEP 14

For $x \geq 0$ , $|x| = x$ .
The integral becomes: \[ \int_{0}^{1} (2 - x - x^2)^2 \, dx $

STEP 15

Calculate each integral separately and add them together to find the total volume.

STEP 16

The volume of the solid is $\frac{16}{3}$ .

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QuestionQ9. Find the volume of the solid object having base in the $x y$ -plane bounded by the curves $y=x^{2}-1$ and $y=1-|x|$ (see image below) and having cross-sections perpendicular to the $x$ -axis that are squares.

Studdy Solution

STEP 1

What is this asking? Find the volume of a 3D shape with a base between two curves, where slices perpendicular to the $x$ -axis are squares. Watch out! Don't mix up the curves; make sure you know which is on top and which is on the bottom!

STEP 2

1. Find the intersection points
2. Determine the side length of the square
3. Set up the integral for volume
4. Calculate the volume

STEP 3

First, we need to find where the curves intersect.
The curves are $y = x^2 - 1$ and $y = 1 - |x|$ .
Set them equal to each other to find the $x$ -values where they meet.

STEP 4

Let's solve $x^2 - 1 = 1 - |x|$ .
We need to consider two cases for $|x|$ : when $x \geq 0$ and when $x < 0$ .

STEP 5

For $x \geq 0$ , $|x| = x$ .
So, we solve: \[ x^2 - 1 = 1 - x $ Rearrange to: \[ x^2 + x - 2 = 0 $ Factor it: \[ (x + 2)(x - 1) = 0 $ So, \(x = -2\) or \(x = 1\).
Since $x \geq 0$ , we take $x = 1$ .

STEP 6

For $x < 0$ , $|x| = -x$ .
So, we solve: \[ x^2 - 1 = 1 + x $ Rearrange to: \[ x^2 - x - 2 = 0 $ Factor it: \[ (x - 2)(x + 1) = 0 $ So, \(x = 2\) or \(x = -1\).
Since $x < 0$ , we take $x = -1$ .

STEP 7

The intersection points are $x = -1$ and $x = 1$ .

STEP 8

The side length of the square is the distance between the curves at any $x$ .
The top curve is $y = 1 - |x|$ and the bottom curve is $y = x^2 - 1$ .

STEP 9

The side length $s$ is: \[ s = (1 - |x|) - (x^2 - 1) $ Simplify it: \[ s = 2 - |x| - x^2 $

STEP 10

The volume $V$ of the solid is the integral of the area of the square cross-sections from $x = -1$ to $x = 1$ .

STEP 11

The area of each square is $s^2$ .
So, the volume is: \[ V = \int_{-1}^{1} (2 - |x| - x^2)^2 \, dx $

STEP 12

Let's break the integral into two parts because of the absolute value: from $-1$ to $0$ and from $0$ to $1$ .

STEP 13

For $x < 0$ , $|x| = -x$ .
The integral becomes: \[ \int_{-1}^{0} (2 + x - x^2)^2 \, dx $

STEP 14

For $x \geq 0$ , $|x| = x$ .
The integral becomes: \[ \int_{0}^{1} (2 - x - x^2)^2 \, dx $

STEP 15

Calculate each integral separately and add them together to find the total volume.

STEP 16

The volume of the solid is $\frac{16}{3}$ .

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QuestionQ9. Find the volume of the solid object having base in the xyx yxy-plane bounded by the curves y=x2−1y=x^{2}-1y=x2−1 and y=1−∣x∣y=1-|x|y=1−∣x∣ (see image below) and having cross-sections perpendicular to the xxx-axis that are squares.

Studdy Solution

STEP 1

What is this asking? Find the volume of a 3D shape with a base between two curves, where slices perpendicular to the xxx-axis are squares. Watch out! Don't mix up the curves; make sure you know which is on top and which is on the bottom!

STEP 2

1. Find the intersection points2. Determine the side length of the square3. Set up the integral for volume4. Calculate the volume

STEP 3

First, we need to find where the curves intersect.The curves are y=x2−1y = x^2 - 1y=x2−1 and y=1−∣x∣y = 1 - |x|y=1−∣x∣.Set them equal to each other to find the xxx-values where they meet.

STEP 4

Let's solve x2−1=1−∣x∣x^2 - 1 = 1 - |x|x2−1=1−∣x∣.We need to consider two cases for ∣x∣|x|∣x∣: when x≥0x \geq 0x≥0 and when x<0x < 0x<0.

STEP 5

For x≥0x \geq 0x≥0, ∣x∣=x|x| = x∣x∣=x.So, we solve: \[ x^2 - 1 = 1 - x $ Rearrange to: \[ x^2 + x - 2 = 0 $ Factor it: \[ (x + 2)(x - 1) = 0 $ So, \(x = -2\) or \(x = 1\).Since x≥0x \geq 0x≥0, we take x=1x = 1x=1.

STEP 6

For x<0x < 0x<0, ∣x∣=−x|x| = -x∣x∣=−x.So, we solve: \[ x^2 - 1 = 1 + x $ Rearrange to: \[ x^2 - x - 2 = 0 $ Factor it: \[ (x - 2)(x + 1) = 0 $ So, \(x = 2\) or \(x = -1\).Since x<0x < 0x<0, we take x=−1x = -1x=−1.

STEP 7

The intersection points are x=−1x = -1x=−1 and x=1x = 1x=1.

STEP 8

The side length of the square is the distance between the curves at any xxx.The top curve is y=1−∣x∣y = 1 - |x|y=1−∣x∣ and the bottom curve is y=x2−1y = x^2 - 1y=x2−1.

STEP 9

The side length sss is: \[ s = (1 - |x|) - (x^2 - 1) $ Simplify it: \[ s = 2 - |x| - x^2 $

STEP 10

The volume VVV of the solid is the integral of the area of the square cross-sections from x=−1x = -1x=−1 to x=1x = 1x=1.

STEP 11

The area of each square is s2s^2s2.So, the volume is: \[ V = \int_{-1}^{1} (2 - |x| - x^2)^2 \, dx $

STEP 12

Let's break the integral into two parts because of the absolute value: from −1-1−1 to 000 and from 000 to 111.

STEP 13

For x<0x < 0x<0, ∣x∣=−x|x| = -x∣x∣=−x.The integral becomes: \[ \int_{-1}^{0} (2 + x - x^2)^2 \, dx $

STEP 14

For x≥0x \geq 0x≥0, ∣x∣=x|x| = x∣x∣=x.The integral becomes: \[ \int_{0}^{1} (2 - x - x^2)^2 \, dx $

STEP 15

Calculate each integral separately and add them together to find the total volume.

STEP 16

The volume of the solid is 163\frac{16}{3}316​.

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QuestionQ9. Find the volume of the solid object having base in the $x y$ -plane bounded by the curves $y=x^{2}-1$ and $y=1-|x|$ (see image below) and having cross-sections perpendicular to the $x$ -axis that are squares.

What is this asking? Find the volume of a 3D shape with a base between two curves, where slices perpendicular to the $x$ -axis are squares. Watch out! Don't mix up the curves; make sure you know which is on top and which is on the bottom!

1. Find the intersection points
2. Determine the side length of the square
3. Set up the integral for volume
4. Calculate the volume

First, we need to find where the curves intersect.
The curves are $y = x^2 - 1$ and $y = 1 - |x|$ .
Set them equal to each other to find the $x$ -values where they meet.

Let's solve $x^2 - 1 = 1 - |x|$ .
We need to consider two cases for $|x|$ : when $x \geq 0$ and when $x < 0$ .

For $x \geq 0$ , $|x| = x$ .
So, we solve: \[ x^2 - 1 = 1 - x $ Rearrange to: \[ x^2 + x - 2 = 0 $ Factor it: \[ (x + 2)(x - 1) = 0 $ So, \(x = -2\) or \(x = 1\).
Since $x \geq 0$ , we take $x = 1$ .

For $x < 0$ , $|x| = -x$ .
So, we solve: \[ x^2 - 1 = 1 + x $ Rearrange to: \[ x^2 - x - 2 = 0 $ Factor it: \[ (x - 2)(x + 1) = 0 $ So, \(x = 2\) or \(x = -1\).
Since $x < 0$ , we take $x = -1$ .

The intersection points are $x = -1$ and $x = 1$ .

The side length of the square is the distance between the curves at any $x$ .
The top curve is $y = 1 - |x|$ and the bottom curve is $y = x^2 - 1$ .

The side length $s$ is: \[ s = (1 - |x|) - (x^2 - 1) $ Simplify it: \[ s = 2 - |x| - x^2 $

The volume $V$ of the solid is the integral of the area of the square cross-sections from $x = -1$ to $x = 1$ .

The area of each square is $s^2$ .
So, the volume is: \[ V = \int_{-1}^{1} (2 - |x| - x^2)^2 \, dx $

Let's break the integral into two parts because of the absolute value: from $-1$ to $0$ and from $0$ to $1$ .

For $x < 0$ , $|x| = -x$ .
The integral becomes: \[ \int_{-1}^{0} (2 + x - x^2)^2 \, dx $

For $x \geq 0$ , $|x| = x$ .
The integral becomes: \[ \int_{0}^{1} (2 - x - x^2)^2 \, dx $

The volume of the solid is $\frac{16}{3}$ .