Volume of a solid, Bounded by a circle (integration)

[Brain0991]

Find the volume of the solid whose base is bounded by the circle x^2+y^2=9 and cross section areas are isosceles right triangles perpendicular to the x-axis?

I am having trouble picturing this figure, i am usually good with geometry type things but i just cannot picture this.
It would be great if someone could help me out with what solid i need to find the volume of and maybe which method to use (shell, washer, etc.)

 

[galactus]

This is NOT rotated. Thus,no shells, washer, etc, method. Use the area of a right isosceles triangle.

The base of the triangle is perp.to the x axis. The length of this base (hypoteneuse) is $\displaystyle y=\sqrt{9-x^{2}}$

Since it is a right isosceles triangle, it's area can be found to be $\displaystyle \frac{y^{2}}{4}$

Integrate from -3 to 3.

$\displaystyle \frac{1}{4}\int_{-3}^{3}y^{2}dx=\frac{1}{4}\int_{-3}^{3}(9-x^{2})dx$

Here's a picture I made in Paint to help show what they mean. I hope it will suffice.

 

[Brain0991]

Wow thank you so much. I definately get it now. *not sarcasm* lol

 

[wjm11]

Find the volume of the solid whose base is bounded by the circle x^2+y^2=9 and cross section areas are isosceles right triangles perpendicular to the x-axis?

Galactus’s explanation is correct, though there is a minor calculation error, I believe.

Base of triangle, b = 2y = 2(9 – x^2)^(.5)
Height of triangle, h = y = (9 – x^2)^(.5)
Area of triangle, A = (1/2)bh = (1/2)( 2(9 – x^2)^(.5))( (9 – x^2)^(.5)) = 9 – x^2 = y^2.

Please check my calculations for accuracy.

 

[galactus]

I do not think there is a calculation error. But, I have been wrong before. :wink:

The base (hypoteneuse lying perp. to x-axis) is length L. Therefore, half the base is L/2.

Since the lower two angles are 45, the height, h, (a bisector drawn up from the middle of the hypoteneuse to the right angle) is L/2 as well.

$\displaystyle \frac{(\frac{L}{2})(\frac{L}{2})}{2}=\frac{L^{2}}{8}$

Since there are two such triangles, $\displaystyle 2\cdot \frac{L^{2}}{8}=\frac{L^{2}}{4}$

Another way. The area of an isosceles is given by $\displaystyle A=\frac{1}{2}s^{2}sin({\theta})$. Where s is the side lengths.

theta is Pi/2, so we have $\displaystyle \frac{1}{2}s^{2}$.

Since the is hypoteneuse is L, The sides have length $\displaystyle s=\sqrt{(\frac{L}{2})^{2}+(\frac{L}{2})^{2}}=\frac{L}{\sqrt{2}}$

$\displaystyle \text{Area}=\frac{1}{2}(\frac{L}{\sqrt{2}})^{2}=\frac{L^{2}}{4}$

Besides that, this is a well known formula for the area of an isosceles in terms of the hypoteneuse.

 

[wjm11]

I agree that A = L^2/4.

However, when we define y as y = (9 - x^2)^(.5), then y = L/2, or L = 2y. Hence, A = (2y)^2/4 = y^2.

 

[BigGlenntheHeavy]

$\displaystyle Base \ of \ triangle \ = \ \sqrt{9-x^2}--\sqrt{9-x^2} \ = \ 2\sqrt{9-x^2}$

$\displaystyle Height \ of \ triangle \ = \ \sqrt{9-x^2}$

$\displaystyle Area \ of \ triangle \ = \ 9-x^2$

$\displaystyle Hence, \ V \ = \ 2\int_{0}^{3}(9-x^2)dx \ = \ 36 \ cu. \ units$

$\displaystyle What \ is \ the \ volume \ if \ the \ triangle \ is \ perpendicular \ to \ the \ y-axis?$