[Math] Volume using cross sections

calculusvolume

Can someone please help me solve this problem?

Q: The base of S is an elliptical region with boundary curve 9x^2 +4y^2 = 36. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

Can someone please help me solve this? How does the sketch of this even look like?

We are trying to solve volume using cross sections, so V(X) = Integral of A(x) from a to b.

What is A(x) in our case, and a, b?

Thanks

Best Answer

enter image description here$$ S : \frac{x^2}{4} + \frac{y^2}{9} = 1 $$ Because the cross sections are isosceles right triangles with hypotenuse in the base, the length of the hypotenuse is $ 6\sqrt{1-\frac{x^2}{4}}$, which means that the area of the cross section is $9(1-\frac{x^2}{4})$

$$ \implies V(x) = \int\limits_{-2}^{2}9(1-\frac{x^2}{4}) \mathrm{d}x $$

That integral should be easy.

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