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

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

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 $\displaystyle x^2 + y^2 = 1$ and the cross sections perpendicular to the $\displaystyle x$-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 $\displaystyle V =\frac{8}{3}$.)

--DragonLord

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

Fernando Revilla

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

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

--DragonLord

4. Originally Posted by DragonLord
Can you give more details on how $\displaystyle b=h=2\sqrt{1-x^2}$ is derived?
For every $\displaystyle x\in [-1,1]$ the correponding ordinates on $\displaystyle x^2+y^2=1$ are $\displaystyle y=\pm \sqrt{1-x^2}$ so, the base of the triangle is $\displaystyle b=\sqrt{1-x^2}-(-\sqrt{1-x^2})=2\sqrt{1-x^2}$ .

Fernando Revilla

5. Thank you!

--DragonLord