Volumes Question (1 Viewer)

Axio · Dec 15, 2014

Could someone help me with this question? When I form the integral I get: $\sqrt{3}\int_{-2}^{2}\sqrt{1-\frac{x^2}{4}} \ dx$

But in the solutions they don't have a square root on the integrand, I'm not sure what I'm doing wrong

. Thanks.

The base of a particular solid is the ellipse: $\frac{x^2}{4}+y^2=1$

Find the volume of the solid if every cross-section perpendicular to the major axis is an equilateral triangle with one side in the base of the solid.

Drongoski · Dec 15, 2014

I think, the cross-section area at x =

$\frac {1}{2} ab sinC = \frac {1}{2} \times 2\sqrt{1-\frac{x^2}{4}} \times 2\sqrt{ 1-\frac {x^2}{4}}\times \frac {\sqrt 3}{2} = (1-\frac{x^2}{4})\sqrt 3$

Axio · Dec 15, 2014

Drongoski said:
I think, the cross-section area at x =

$\frac {1}{2} ab sinC = \frac {1}{2} \times 2\sqrt{1-\frac{x^2}{4}} \times 2\sqrt{ 1-\frac {x^2}{4}}\times \frac {\sqrt 3}{2}$

Ok thanks. I think I know where I stuffed up.

Volumes Question (1 Viewer)

Axio

=o

Drongoski

Well-Known Member

Axio

=o

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