# Thread: Find the volume of the solid

1. ## Find the volume of the solid

Hi,

I'm having trouble with this question as I'm sure how the do the problem when it involves the isosceles triangle. I be great if someone could possibly draw me a diagram and show me how to form the equation to use and evaluate the definite integral that is obtained from the intercepts that I already know to be the square root of 1/2 and -1/2.

Thanks,

Hi,

I'm having trouble with this question as I'm sure how the do the problem when it involves the isosceles triangle. I be great if someone could possibly draw me a diagram and show me how to form the equation to use and evaluate the definite integral that is obtained from the intercepts that I already know to be the square root of 1/2 and -1/2.

Thanks,
The base of each isosceles triangle is $\displaystyle x^2-(1-x^2)=2x^2-1$.

Since the height of each isosceles triangle is the same as its base, then for any $\displaystyle x\in\left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$, the area of each slice is $\displaystyle A(x)=\frac{1}{2}\left(2x^2-1\right)^2$.

So to find the volume, evaluate $\displaystyle V=\frac{1}{2}\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}(2x^2-1)^2\,dx=\int_{0}^{\frac{1}{\sqrt{2}}}(2x^2-1)^2\,dx$

Can you continue?

3. Thanks a lot Chris L T521.

I managed to find A(x) but I found I made a mistake as I cancelled the ½ and the square for some reason so I only had A(x) as (2x^2-1)

I then found V but I just wasn't sure that it was equal to the other definite integral.

I found the answer 0.3771 to be the same evaluating either integral so I assume this is correct?

That would be correct. The exact answer would be preferred: $\displaystyle \frac{4}{15}\sqrt{2}$