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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
6장, 문제 6.4.5c

Let R be the region in the first quadrant bounded above by the curve y=2−x² and bounded below by the line y=x. Suppose the shell method is used to determine the volume of the solid generated by revolving R about the y-axis.
Illustration of the shell method for calculating volume, showing region R, shell height, and radius in the first quadrant.
c. Write an integral for the volume of the solid using the shell method.

검증된 단계별 안내
1
Identify the region R bounded by the curves y = 2 - x^2 (above) and y = x (below) in the first quadrant. This region is revolved around the y-axis to form the solid.
Recall that the shell method involves integrating with respect to x when revolving around the y-axis. Each shell has a radius equal to the distance from the y-axis, which is x, and a height equal to the difference between the upper and lower functions: height = (2 - x^2) - x.
Write the volume of a typical shell as the circumference times the height times the thickness: Volume of shell = 2\(\pi\) \(\times\) (radius) \(\times\) (height) \(\times\) (thickness) = 2\(\pi\) x \(\big\)((2 - x^2) - x\(\big\)) \, dx.
Determine the limits of integration by finding the x-values where the two curves intersect in the first quadrant. Solve 2 - x^2 = x to find these points.
Set up the integral for the volume as: \(V = \int_{a}^{b} 2\pi x \big((2 - x^2) - x\big) \, dx\), where a and b are the intersection points found in the previous step.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
6m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Shell Method for Volume

The shell method calculates the volume of a solid of revolution by summing cylindrical shells. Each shell's volume is approximated by 2π(radius)(height)(thickness), where the radius is the distance from the axis of rotation, the height is the function value difference, and the thickness is a small change in the variable of integration.
추천 영상:
04:48
Finding Volume Using Disks

Setting up the Integral with Respect to x

When revolving around the y-axis, the shell radius is the x-value of the shell, and the height is the difference between the upper and lower functions, here y=2−x² and y=x. The integral sums shells from the leftmost to rightmost x-values defining the region, integrating with respect to x.
추천 영상:
03:39
Integrals of Natural Exponential Functions (e^x)

Determining the Bounds of Integration

The bounds are found by identifying where the curves intersect in the first quadrant. Solving y=2−x² and y=x gives the limits for x, which define the interval over which the shells are integrated to find the volume.
추천 영상:
05:06
Finding Area When Bounds Are Not Given
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