Determining VOLUME via Cross-Sections! 1) Use the general...

Determining VOLUME via Cross-Sections! 1) Use the general slicing method to find the volume of the following solid. The solid whose base is the region bounded by the curves $x = y^2$ and $y = \pm \sqrt{2x - 4}$, and whose cross-sections through the solid perpendicular to the x-axis are right-isosceles triangles. The right angle is at the peak (NOT in the foot-print region), and the triangles are perpendicular to the y-axis.

Transcript

00:01 So in this question, they say, i want to use the general slicing method to find the volume of the solid.

00:06 The solid is the one whose base is the region that's bounded by the curve, y equals 12 times the squared of cosine x, and the x -axis from negative pi over 2 to positive pi over 2.

00:16 And whose cross -sections, through the solid, perpendicular to the x -axis, are isosceles right triangles with a horizontal leg in the x -y plane and a vertical leg above the x -axis.

00:28 So let's start by sketching the region that's in the xy plane.

00:33 I have this region that is bounded above by y equals 12 times the square root of cosine x.

00:44 And i need to know what are my x intercepts.

00:47 So my x intercepts are when y is zero, so when 0 equals 12 square root of cosine x, when 0 equals square root cosine x, when 0 equals square root cosine x, when 0 equals cosine x, so x equals negative pi over 2 or positive pi over 2.

01:10 Now, perpendicular to the x -axis over top of this region, i am building these isosceles -right triangles.

01:21 And so the question is, how do i find the volume of a cross -sectional zon? i have a formula, and that formula is volume equals the integral from a to b of the area of a cross -section as a function of x dx.

01:38 Now, in this case, my region extends from x equals negative pi over 2 to positive pi or 2.

01:47 Now, what's the area of a cross -section? the area of an isosceles right triangle.

01:53 Now, the area of a triangle is 1 half base times height.

01:57 But in an isosceles right triangle, my base and my height are the same...

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