๐Ÿ“š

All Subjects

ย >ย 

โ™พ๏ธย 

AP Calc

ย >ย 

๐Ÿถ

Unit 8

8.7 Volumes with Cross Sections: Squares and Rectangles

2 min readโ€ขjune 8, 2020

Anusha Tekumulla


Finding Volume Using Square Cross Sections

Now that you can find the area between curves, you should have no problem finding the volumes of 3-dimensional shapes. In order to find the area of a shape using a square cross section, take a look at the example below.ย 

๐Ÿ” Example Problem: Finding the Volumes of a Shape with Square Cross Sectionsย ย 

Letโ€™s say we have a function x = 4 - y^2 and we know that the cross sections are squares. Because the height and width of squares are the same, finding the volume is quite easy. Before, we were finding the area by adding up a bunch of infinitely thin slices. Now, we are finding the volume by adding up a bunch of infinitely thin squares.

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2FScreenshot%20(662).png?alt=media&token=50c7b772-8156-4ae6-a746-c62dc1842ac0

The area of one square cross section is ((4 - y^2) - -(4 - y^2))^2. This can be simplified to (2(4 - y^2))^2. This is a simple right minus left represented in topic 8.5. Now we integrate from the intersection points found by setting y = 0.

Using Rectangular Cross Sections

Like before with squares, we can use rectangles to find the volume of an object. The area of a rectangle is A = lw, where l is length and w is width. In each case, once we figure out l and w, we can integrate between our bounds to find the volume.

๐Ÿ” Example Problem: Finding the Volumes of a Shape with Rectangular Cross Sections

https://firebasestorage.googleapis.com/v0/b/fiveable-92889.appspot.com/o/images%2Fb3635f10d6d9bbec1dad58f1fcebbed4.png?alt=media&token=270190f6-ddb7-4751-a28c-8030c0aba4ed

Problem Courtesy of KhanAcademy

We're given the function y = x^3/2 and told to find a volume with rectangular cross sections perpendicular to the y-axis. This is an important distinction. This means that the length of each rectangle is NOT x^3/2 (y), but rather (2y)^(1/3) [this is obtained by solving for x], since our rectangles are in terms of y, not x. Now, we just need to find l. Luckily, we're simply given this in the problem. Therefore, our integral will be the integral from 0 to 4 of (2y)^(1/3) * y/2.

Was this guide helpful?

Join us on Discord

Thousands of students are studying with us for the AP Calculus AB/BC exam.

join now

Browse Study Guides By Unit

โœ๏ธ
Free Response Questions (FRQ)

๐Ÿง
Multiple Choice Questions (MCQ)

โ™พ
Unit 10: Infinite Sequences and Series (BC Only)

๐Ÿ‘‘
Unit 1: Limits & Continuity

๐Ÿค“
Unit 2: Differentiation: Definition & Fundamental Properties

๐Ÿค™๐Ÿฝ
Unit 3: Differentiation: Composite, Implicit & Inverse Functions

๐Ÿ‘€
Unit 4: Contextual Applications of the Differentiation

โœจ
Unit 5: Analytical Applications of Differentiation

๐Ÿ”ฅ
Unit 6: Integration and Accumulation of Change

๐Ÿ’Ž
Unit 7: Differential Equations

๐Ÿฆ–
Unit 9: Parametric Equations, Polar Coordinates & Vector Valued Functions (BC Only)

Play this on HyperTyper

Practice your typing skills while reading Volumes with Cross Sections: Squares and Rectangles

Start Game