4599: WeBWorK Logged in MATHEMATICAL ASSOCIATION OF AMERICA webwork/m211dabrowsk

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4599: WeBWorK Logged in MATHEMATICAL ASSOCIATION OF AMERICA webwork/m211dabrowskisummer16/6b 3 6b: Problem 3 Prev Up (1 p) Find the volume of the solid Sthat satisfies the two following conditions. First, the base of S is the elliptical region with boundary curve 9x 4y 64, and second, the cross-sections of S perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base. volume of S Preview Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. Ermal Instructor &showold; Answers

Explanation / Answer

Solution:

9x² + 4y² = 64
y = ±(1/2) (64 - 9x²)

So, we have the two curves:
y = (-1/2)(64 - 9x^2) and y = (1/2)(64 - 9x^2).

The distance between these two curves is:
(1/2)(64 - 9x^2) - (-1/2)(64 - 9x^2) = (64 - 9x^2),

so this is the length of the hypotenuse.

Now, call s the length of the two legs of the isosceles right triangle. Then, by the Pythagorean Theorem:

s^2 + s^2 = [(64 - 9x^2)]^2 = 64 - 9x^2 ==> s^2 = (1/2)(64 - 9x^2).

The area of the isosceles triangle is:
A = (1/2)bh = (1/2)(s)(s) = (1/2)s^2 = (1/4)(64 - 9x^2).

To obtain the volume of the solid, we need to integrate the formula for the area of the cross-section over the x-values of the region (-2 x 2). Therefore, the required volume of S is:

V(S) = (1/4)(64 - 9x^2) dx (from x=-2 to 2)
= 1/4 (64 - 9x^2) dx (from x=-2 to 2)
= (1/4)(64x - 3x^3) (evaluated from x=-2 to 2)
= (1/4)[(128 - 24) - (-128 + 24)]
= 52 units^3.

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