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In this problem, we will find the volume of a solid with circular base of radius 2, for which parallel cross-sections perpendicular to the base are squares. To do this, we will assume that the base is the circle x2+y2=4, so that the solid lies between planes parallel to the x-axis at x=2 and x=−2. The cross-sections perpendicular to the x-axis are then squares whose bases run from the semicircle y=√−4−x2 to the semicircle y = √4−x2

Required:
a. What is the area of the cross-section at x?
b. What is the volume of the solid ?

Respuesta :

LammettHash

(a) Each cross section is a square whose side length is decided by the distance in the x,y plane between the two curves [tex]y=\sqrt{4-x^2}[/tex] and [tex]y=-\sqrt{4-x^2}[/tex], which is [tex]2\sqrt{4-x^2}[/tex]. Then each cross section has area

[tex](2\sqrt{4-x^2})^2=4(4-x^2)=\boxed{16-4x^2}[/tex]

(b) The volume of the solid is obtained by integrating the cross-sectional area from x = -2 to x = 2.

[tex]\displaystyle\int_{-2}^2(16-4x^2)\,\mathrm dx=16x-\dfrac43x^3\bigg|_{-2}^2=\boxed{\dfrac{128}3}[/tex]

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