The base of a solid V is the circle x^2+y^2=1 . Find the volume if V has (a) square cross sectio

The base of a solid V is the circle x^2+y^2=1 . Find the...

Key Concepts

Volume by Slicing Method

This technique involves dividing a solid into many thin cross-sectional slices perpendicular to a given axis, calculating the area of each slice as a function of the position along the axis, and then integrating these areas over the range of the solid. It is a fundamental approach in calculus for finding the volume of solids with known cross-sectional shapes.

Cross-Sectional Area Calculation

For any solid whose cross sections have a specific geometric shape, determining the area of each slice is essential. This involves expressing the dimensions of the cross section in terms of the position along the axis of integration and applying the appropriate formula for the area of that shape, whether it is a square, a semicircle, or another figure.

Geometric Representation and Determining Integration Limits

In problems where the base of a solid is defined by a geometric figure, such as a circle, understanding the relationship between the shape’s equation and the dimensions of the cross sections is key. This allows for the accurate determination of integration limits and expressions for the cross-sectional dimensions as functions of the position variable.

Transcript

00:01 Okay, i think the toughest thing for these problems is realizing that you have two different functions that you need to do.

00:11 The semicircle that's on top and a semicircle that's on bottom.

00:15 What's nice, though, is the formula is pretty easy that you're going from negative one to one, and then you have to take your top function minus your bottom function.

00:28 And it's in terms of x because it's perpendicular to the x axis.

00:32 And what you're really doing here is you're finding the length of a side length, and you're finding each individual square, and then the integral adds them all together for you.

00:46 So what's the upper function? well, if you solve for y in this problem, you would first want to subtract x squared over, but then you would square root to solve for it.

00:57 Now, the positive one will be the upper one, and y equals negative square root.

01:03 Because don't forget when you square root, you get two answers, positive and negative.

01:06 So the positive is the above semicircle.

01:08 The negative is going to be the below semicircle.

01:11 So what's nice about this, and you might just need to go to a calculator, is your upper function of 1 minus x squared, and then you subtract off the lower function, negative square root of 1 minus x squared.

01:29 This is really the same thing.

01:31 Let me just write it over here because it's starting to get busy.

01:34 It's adding, because when you subtract a negative, you add.

01:38 So there's really, from negative one to one, two of these square root functions that need to be squared.

01:47 So you can cancel out that square root with this as long as you remember to square this and then distribute that four in.

01:56 What we're really looking at integral from negative 1 to 1 of the term 4 minus 4x squared.

02:05 It's 2 squared is 4.

02:06 4 times 1 is 4, 4 times x squared is 4x squared, dx.

02:10 And then we can do this math for x minus 4 thirds, x cubed, and that's from negative 1 to 1.

02:19 And when you plug in 1, 4 times 1 is 4 minus 4 thirds because 1 cubed is still 1...

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