Problem

Solve Example 9 taking cross-sections to be paral…

06:14

Problem 63 Hard Difficulty

(a) Set up an integral for the volume of a solid torus (the donut-shaped solid shown in the figure) with radii $ r $ and $ R $.
(b) By interpreting the integral as an area, find the volume of the torus.

Answer

a) 8$\pi R \int_{0}^{r} \sqrt{r^{2}-y^{2}} d y$
b) $V=2 \pi^{2} r^{2} R$

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Video Transcript

So for this problem, Uh, we're looking for volume, and it's the volume found by rotating a circle around the y axis. Um, so this is going to give us what is known as a Taurus? Um, and we'll have X minus the radius squared plus y squared equals r squared. And what will end up using is the washer method to solve for X. So when we saw for X, we get that X is equal to R plus or minus two square root of R squared minus y squared. We'll have the positive being the right half and the negative being the left half so we can use symmetry. We know at least symmetrical. So we will integrate from zero to our and just double it. We know that our radius one is going to be the positive, and our radius two is going to be the negative. So we'll subtract inner from outer, and we get that the volume is equal to two times the integral from zero to our because we doubled it, remember? And that can be times pi or of pi times. The first radius, which we know to be R plus the squared of R squared minus y squared and that value is going to be squared minus. So the outer ar minus the square root of R squared minus y squared and that value squared. We integrate with respect to why and what we end up getting when we simplify is eight pi r times in a row from zero to our of r squared minus y squared dy then for part B, we're going to make use of solving. When we saw for excellent square both sides, we get the equation of a circle. So what we know is that X is representing the right half in quadrants. One in four, um, the integral limits zero to our, uh so that's half a circle, which is 1/4 it's half of the half circle, so it's 1/4 of the full circle. So what we can do is replace the integral with the area of a circle, which we know to be pi r squared and then multiply that by 1/4. So we're going to plug that in here, and what we end up getting is that this is going to be eight pi r times 1/4 pi r squared. So that'll give us our final answer of two pi squared r squared big are

Carson M.

California Baptist University

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Calculus: Early Transcendentals

Chapter 6

Applications of Integration

Section 2

Volumes

Problem

Solve Example 9 taking cross-sections to be paral…

Problem 63 Hard Difficulty

(a) Set up an integral for the volume of a solid torus (the donut-shaped solid shown in the figure) with radii $ r $ and $ R $.
(b) By interpreting the integral as an area, find the volume of the torus.

Answer

a) 8$\pi R \int_{0}^{r} \sqrt{r^{2}-y^{2}} d y$
b) $V=2 \pi^{2} r^{2} R$

More Answers

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Calculus: Early Transcendentals

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Video Transcript

Calculus: Early Transcendentals

Heather Z.

Caleb E.

Kristen K.

Michael J.

Integration Review - Intro

Definite vs Indefinite

a) 8$\pi R \int_{0}^{r} \sqrt{r^{2}-y^{2}} d y$b) $V=2 \pi^{2} r^{2} R$

Calculus: Early Transcendentals

Catherine R.

Heather Z.

Kristen K.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Heather Z.

Caleb E.

Kristen K.

Michael J.

a) 8$\pi R \int_{0}^{r} \sqrt{r^{2}-y^{2}} d y$
b) $V=2 \pi^{2} r^{2} R$