Volumes
Find the volumes of the solids in Exercises 1–18.
The solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. The cross-sections of the solid perpendicular to the x-axis between these planes are circular disks whose diameters run from the curve x² = 4y to the curve y² = 4x.

검증된 단계별 안내

First, identify the region bounded by the curves given: \(x^{2} = 4y\) and \(y^{2} = 4x\). These curves define the limits of the diameter of the circular cross-sections perpendicular to the x-axis between \(x=0\) and \(x=4\).

Rewrite each curve to express \(y\) in terms of \(x\) to find the upper and lower boundaries of the diameter at each \(x\). From \(x^{2} = 4y\), solve for \(y\) to get \(y = \frac{x^{2}}{4}\). From \(y^{2} = 4x\), solve for \(y\) to get \(y = \sqrt{4x} = 2\sqrt{x}\) (considering the positive root since diameters are positive lengths).

At each \(x\) between 0 and 4, the diameter of the circular cross-section is the vertical distance between the two curves: \(D(x) = 2\sqrt{x} - \frac{x^{2}}{4}\).

The radius of the circular cross-section is half the diameter: \(r(x) = \frac{D(x)}{2} = \frac{1}{2} \left( 2\sqrt{x} - \frac{x^{2}}{4} \right)\).

The volume of the solid is found by integrating the area of the circular cross-sections along the \(x\)-axis from 0 to 4. The area of each cross-section is \(A(x) = \pi [r(x)]^{2}\). Therefore, the volume is \(V = \int_{0}^{4} \pi \left( \frac{2\sqrt{x} - \frac{x^{2}}{4}}{2} \right)^{2} dx\).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

10m

주요 개념

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

Volume of Solids with Known Cross-Sections

This concept involves finding the volume of a solid by integrating the area of its cross-sections perpendicular to an axis. When cross-sections have a known shape, such as circles, the volume is the integral of the area function along the axis of interest.

Finding the Diameter from Curves

To determine the diameter of each circular cross-section, you must find the vertical distance between the two curves at a given x. This requires solving each curve for y in terms of x and subtracting to get the length of the segment forming the diameter.

Setting up and Evaluating Definite Integrals

Once the area of each cross-section is expressed as a function of x, the volume is found by integrating this area from the lower to upper bounds (x=0 to x=4). Proper setup of the integral and accurate evaluation are essential to find the exact volume.