Volume of solid given base and cross sections - trouble understanding question

**DragonLord** · February 12th 2011, 06:20 PM

The following is an exercise from Rogawski's Calculus: Early Transcendentals, section 6.2:

In Exercises 9-14, find the volume of the solid with given base and cross sections.

9. The base is the unit circle $x^2 + y^2 = 1$ and the cross sections perpendicular to the $x$ -axis are triangles whose height and base are equal.

I don't clearly understand the question. What exactly are the cross sections ("perpendicular" is ambiguous), and what do I need to integrate? (The answer key says $V =\frac{8}{3}$ .)

--DragonLord

**FernandoRevilla** · February 13th 2011, 01:24 AM

For $x\in[-1,1]$ the base and the height of the triangle are $b=h=2\sqrt{1-x^2}$ so, the area is $A(x)=2(1-x^2)$ . Then, $V=\int_{-1}^{1}A(x)\;dx=\ldots=8/3$

Fernando Revilla

**DragonLord** · February 13th 2011, 05:35 AM

Originally Posted by FernandoRevilla

For $x\in[-1,1]$ the base and the height of the triangle are $b=h=2\sqrt{1-x^2}$ so, the area is $A(x)=2(1-x^2)$ . Then, $V=\int_{-1}^{1}A(x)\;dx=\ldots=8/3$

Fernando Revilla

Can you give more details on how $b=h=2\sqrt{1-x^2}$ is derived? If possible, post a diagram so I'll understand better. Thanks,

--DragonLord

**FernandoRevilla** · February 13th 2011, 10:24 AM

Originally Posted by DragonLord

Can you give more details on how $b=h=2\sqrt{1-x^2}$ is derived?

For every $x\in [-1,1]$ the correponding ordinates on $x^2+y^2=1$ are $y=\pm \sqrt{1-x^2}$ so, the base of the triangle is $b=\sqrt{1-x^2}-(-\sqrt{1-x^2})=2\sqrt{1-x^2}$ .

Fernando Revilla

**DragonLord** · February 13th 2011, 11:36 AM

Thank you!

--DragonLord

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