# calculus

#3 A solid has a base in the form of the ellipse: x^2/25 + y^2/16 = 1. Find the volume if every cross section perpendicular to the x-axis is an isosceles triangle whose altitude is 6 inches.

#4 Use the same base and cross sections as #3, but change the axis to the y-axis.

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1. using symmetry, we can see that since the area of each triangle is 1/2 yz

v = 2∫[0,5] yz dx
where y = 4√(1-x^2/25) and z=6
v = 48/5∫[0,5] ∫(25-x^2) dx = 800

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2. Let's just do the quarter of it over the right half 0 < x < 5 and multiply by 2 at the end
that means the triangle has base 2y and height 6 everywhere so its Area is (1/2)(2y)(6) = 6y and a slice of volume is 6ydx
we need to integrate that from x = 0 to x = 5 (and not forget to double the answer)
y^2 = 16(1 - x^2/25)
y = 4 (1 - x^2/25)^.5 (just use + root)
so we have
24 integral(1 - x^2/25)^.5 dx
or
(24/5) integral (25 - x^2)^.5 dx
=12/5 [ x sqrt(25-x^2)+25sin^-1(x/5) ] }
(looked up sqrt(p^2-x^2) dx)
from x = 0 to x = 5
at x = 5
12/5 [ 0 + 25 sin^-1(1) ]= 12/5[25 pi/2]
=30 pi
at x = 0
12/5[ 0 + 0] = 0
so
30 pi now times 2 = 60 pi
for heavens sake check my arithmetic

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3. It is an ellipse on the bottom, pi has to be involved in the volume I think.

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4. Ah, yes. I forgot the √ in my integral.

I also get 60π

I thought it strange that there was no π, but I was in a hurry at the time.

Score one more for Damon.

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5. A simpler solution uses the fact that the area of an ellipse is πab

Here, a=4 and b=5, so the area of the ellipse is 20π.

Since all the triangles have the same height (6), the volume is (1/2)(6)(20π) = 60π

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