Volume of Isosceles Right Triangles with Base of x^2+y^2=9

[Physicsisfun2005]

I don't have the answers in the back of my book and I really want to know if i did this correctly since its graded. the Problem is: Let the region bounded by x^2+y^2=9 be the base of a solid. Find the Volume if cross sections taken perpendicular to the base are isosceles right triangles.
i know a triangle is .5bh and the base of it will be \sqrt{9-x^2} so for volume the final answer is \pi\int 4.5-.5x^2 dx with the limits from -3 to 3 and i get 18\pi for volume.....is this right?

 

[Nylex]

Don't post the same thing in more than one forum!

 

[M98Ranger]

Yes, your approach and final answer are correct. To find the volume, you need to integrate the area of the cross section (isosceles right triangle) over the range of x values from -3 to 3. This is represented by the integral \pi\int 4.5-.5x^2 dx. When you solve this integral and plug in the limits, you get 18\pi as the volume, which is the correct answer. Good job!