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Ch. 8 – Applications of Definite Integrals 8.3 – Volumes.

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Presentation on theme: "Ch. 8 – Applications of Definite Integrals 8.3 – Volumes."— Presentation transcript:

1 Ch. 8 – Applications of Definite Integrals 8.3 – Volumes

2 The volume of a solid with a known, integrable cross-sectional area of A(x) from x=a to x=b is Ex: Find the volume of the solid formed by revolving the region of the curve between the x-axis and y = 2 + cosx around the x-axis over [0, 2π]. –Graph the function from [0, 2π]. Picture the shape visually. –The revolution of this shape will produce an hourglass-like shape –The formula above says we need a cross-sectional area formula, and since the cross-section is a circle, here’s our formula: –Now integrate over the proper interval. Use your calculator. r

3 Ex: The region enclosed by the y-axis and the graphs of y = x + 2 and y = 4 – x 2 is revolved around the x-axis to form a solid. Find its volume. –Find a cross-sectional area formula: –What are the limits of integration? –Now integrate over the proper interval. Use your calculator. r1r1 r2r2 This is called the washer method because the cross- section is a washer

4 Ex: The region enclosed by the graphs of and y = x + 3 is revolved around the x-axis to form a solid. Find its volume. NO CALCULATOR! –Find a cross-sectional area formula: –What are the limits of integration? –Now integrate over the proper interval. r1r1 r2r2

5 Ex: The region enclosed by the x-axis and the graph of is revolved around the line y = 3 to form a solid. Find its volume. NO CALCULATOR! –Use the washer method, but the washer radii will be trickier to find: –Whew! Limits of integration: –Now integrate over the proper interval.

6 Ex: The region enclosed by the x-axis and the graph of over [1, 4] is revolved around the y-axis to form a solid. Find its volume. –When the radius of the cross-section is parallel to the axis of revolution, use the shell method! r(x) is the radius as a function of x h(x) is the height as a function of x –Find radius and height formulas: –Now integrate over the proper interval.

7 Ex: The region enclosed by the line x = 1, the x-axis, and the graph of is revolved around the line x = 3 to form a solid. Find its volume. NO CALCULATOR! –Use the shell method, but watch out for the tricky radius! –Find radius and height formulas: –Now integrate over the proper interval. x3 – x

8 Ex: A region is bounded by the x-axis, x = 3, and. This region is cut into square cross-sections that sit perpendicular to the x-axis. Find the volume of the solid generated by these cross- sections without a calculator. –I’ll cheat and let you look at my calculator for the graph. –We need a cross-sectional area formula for a square: –Now integrate over the proper interval.

9 Ex: A region is bounded by the x-axis, x = 3, and. This region is cut into equilateral triangle cross-sections that sit perpendicular to the x-axis. Find the volume of the solid generated by these cross-sections without a calculator. –I’ll cheat and let you look at my calculator for the graph. –Know this formula for equilateral triangle cross-sections: –Now integrate over the proper interval.


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