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Problem 41 Hard Difficulty

A torus is generated by rotating the circle $ x^2 + (y - R)^2 = r^2 $ about the x-axis. Find the volume enclosed by the torus.


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 3

Trigonometric Substitution

Related Topics

Integration Techniques

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Tessa S.

September 18, 2022

Use cylindrical shells to find the volume of the torus obtained by revolving the circle ? ? + ? ? = ? ? about the line ? = ?, where ? > ? > 0.

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01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

Video Thumbnail

27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Watch More Solved Questions in Chapter 7

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Problem 9
Problem 10
Problem 11
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Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
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Problem 24
Problem 25
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Problem 27
Problem 28
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Problem 31
Problem 32
Problem 33
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Problem 44

Video Transcript

a tourist is generated by rotating the circle X squared plus y minus capital are square is small, are square So that's this blue circle up here and conference one into we'LL rotate that about the X axis And down here we have another circle and the volume will weaken Take a cross section here and because of this large gap in the middle, we could see that the cross sections or washers And we we could obtain this volume by letting the washer trace out this volume in the ex direction where since the circle has really is our we have negative are over here the X value are over here. So these are our bounds of integration, Rex. And now for the washer We know that we'LL want usually the general formula for washers, pie and then large radius All due notice by are big square minus are small So this is it for using d X. So these Andy or the X pounds of integration Now we'LL have to go ahead and find big radius and small radius. So looking at the figure here we see that we could the radius Well, here's the center of the washer. Here's the origin in and to find the Big Radius it looks like we go up here, Toe are and then we have to add in that semi circle. So here's one way to find are big So let's look at this and my circle, but placed at the origin so that will be explored. Plus, y square equals R square And then we'LL go ahead and shift up the circle by our units. So after we do that, we shifted up by Are the circle looks like this. So in that case, we see that the total distance it's just are added to this upper half of the semi circle which is why equals from this equation right here from the circle R squared minus explode So are big We'll just be are the shift plus the upper half of the circle it's being radius and then similarly for the small radius are small Same idea but we'LL look at the lower half of the circle So this time we'Ll take that lower half circle and then we'LL shift it up I r and this will give us the smaller radius over here and dotage That's a small radius. So this time since we're on the negative circle our small B the shift up that we added in the vertical direction that's our And in this time it's plus. But we're doing the negative part of the circles of this time. It's negative R squared, minus explorers. So this is the equation for this Lower half someone circle exes. So this equation was the negative are slurred minus X squared. So we have found our big in our small, So we'LL go ahead and plug those into our volume formula. Let's go to the next page for that. We have v picking up where we left off and we observed that the bonds were negative. Arts are in the ex direction. We have pie. Let's pull the pie in the front of the girls. Before then, we have our Dade squared, so we have our plus the radical square the whole thing. So our big square and now we'll do our small squares. Then let's just go ahead and evaluate the squares so we have our square to our times the radical plus and then we square the radical. So this is the first square and then for the second square and then plus R squared minus x square. And then let's go ahead and cancel out as much as we can. So we have our square minus R squared. Those cancel r squared minus x square and then with a minus part squared minus X for years. And here the radicals to two are radicals Won't cancel because the double negative. So we get for our something, just go ahead and pull. The four are the constant for big Our pie. Yeah, in a rural negative. RTR and then we're left with Where was that? Excuse me, R squared minus little exclude yet I sees me. I pulled out. The four are here and I'm left with r squared minus x word. And now, with ready for a tricks up. So here you should take extra PR sign. What? Thandie Exes are co signed data. Now here we have to be a little careful. We are dealing with the definite general. So these bounds these limits of integration are subject to change. So to change these, let's go to our tricks up and recall anytime you sign entrance up years your restriction for data. So let's plug in these X values. So the lower bound X is negative arm and then the right hand side is our assigned data. This means signed data is negative one and in December hold on ly time that happens is when data's negative Piper too similarly for the upper bound X equals our plug. It in the left hand side becomes our and the right hand side. Our scientist, This means scientist is one and the only time that happens in our interval is that pirate too. So these air the fate of bounds of integration And let's go ahead and simplify this radical before we go on R squared minus x squared. This is equal to the radical R squared minus our square science where cool it are. A lot are square and then we have a radical still one minus sign square. This is our times a squirt of coastline square and that's just our co signed data so that the radical becomes our coastline data. But also DX gives you another Arcosanti data. So after multiply these arcos on Data's together so that'LL be our next date. Let's go to the next page we have for our pie negative, however, to two pirates who and then we'LL supplying the r co signed and our co sign We have r squared co sign squared data data. Now this little ours is not a variable That data is the variable so we can go ahead and pull the r squared out. And also for this coastline, let's use the half angle identity Coastline square is one plus coastline to data all over too. So pull out the two from the denominator So you have four over to which is too big our pie little r squared and then one plus co sign to data we can integrate this integral of one Just data in a rolls co signed today No scientific data over to and we have our bounds negative power, Hutu power Hutu And I will just put those in Data's pirate too. And then we'LL have sign of pie over Zoom And now we plug in negative however too plus sign of negative pie. Oliver too. We no sign of pie inside of minus fireball zero you have to capital are pie little r squared. Then you have pie or two plus a pie or two. So that's just another pie in there. So you'LL have it times pi And then we could simplify this by just squaring that by multiplying goes together and there's our final answer.

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Video Thumbnail

01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

Video Thumbnail

27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

Join Course
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