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

In summary, the conversation discusses finding the volume of a solid with a base bounded by x^2+y^2=9, where the cross sections are isosceles right triangles. The formula used is .5bh, with the base being \sqrt{9-x^2}. The final answer for volume is \pi\int 4.5-.5x^2 dx, with limits from -3 to 3, resulting in a volume of 18\pi. The individual is asking for confirmation that their answer is correct.
  • #1
Physicsisfun2005
70
0
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 [tex]x^2+y^2=9[/tex] 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 [tex].5bh[/tex] and the base of it will be [tex]\sqrt{9-x^2}[/tex] so for volume the final answer is [tex]\pi\int 4.5-.5x^2 dx[/tex] with the limits from -3 to 3 and i get [tex]18\pi[/tex] for volume.....is this right?
 
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  • #2
Don't post the same thing in more than one forum!
 
  • #3


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!
 

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

1. What is the formula for finding the volume of an isosceles right triangle with a base of x^2+y^2=9?

The formula for finding the volume of an isosceles right triangle with a base of x^2+y^2=9 is V = (1/3) * (x^2 + y^2) * h, where h is the height of the triangle.

2. How do you find the height of an isosceles right triangle with a base of x^2+y^2=9?

To find the height of an isosceles right triangle with a base of x^2+y^2=9, you can use the Pythagorean Theorem. Since the triangle is isosceles, the two equal sides are the legs of a right triangle. Therefore, the height can be found by taking the square root of (x^2 + y^2)/2.

3. Can the volume of an isosceles right triangle with a base of x^2+y^2=9 be negative?

No, the volume of any geometric shape cannot be negative. It represents the amount of space occupied by the shape and is always a positive value.

4. Does the value of x and y affect the volume of an isosceles right triangle with a base of x^2+y^2=9?

Yes, the value of x and y will affect the volume of the isosceles right triangle. As the value of x and y changes, the length of the base and height of the triangle will also change, resulting in a different volume.

5. Can the volume of an isosceles right triangle with a base of x^2+y^2=9 be larger than the volume of a cube with the same base length?

It is not possible for the volume of an isosceles right triangle to be larger than the volume of a cube with the same base length. A cube has a larger volume than any other geometric shape with the same base length. Therefore, the volume of the isosceles right triangle will always be smaller than that of the cube.

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