Volume of solid having triangle base and semicircle slice

[songoku]

Homework Statement

Calculate the volume of solid having triangle base with vertices (0, 0) , (2, 0) and (0, 3) whose slice perpendicular to x-axis is semicircle

Relevant Equations

Volume = $\int_p^{q} A(x) dx$

First, I tried to find the equation of line passing through (2, 0) and (0, 3) and I got $y=3-\frac{3}{2}x$

Then I set up equation for the area of one slice, $A(x)$
$$A(x)=\frac{1}{2} \pi r^2$$
$$=\frac{1}{2} \pi \left( \frac{1}{2}y\right)^2$$
$$=\frac{1}{2} \pi \left(\frac{3}{2}-\frac{3}{4}x \right)^2$$

Am I correct until this point? Thanks

 

[anuttarasammyak]

I am not sure I got right image of the solid. Could you show me your sketch ?

 

[songoku]

I am not sure I got right image of the solid. Could you show me your sketch ?

Hopefully it is clear enough

Thanks

 

[anuttarasammyak]

Thanks. Clear enough. So we observe the volume of the solid is half the volume of the cone which has base of area $(3/2)^2\pi$ and height 2. It would be beneficial for you to check answer.

Your nice sketch leads me to
$$V=\frac{1}{2}\int_0^2 \pi [\frac{3- \frac{3}{2}x}{2}]^2 dx$$

 

[songoku]

Thanks. Clear enough. So we observe the volume of the solid is half the volume of the cone which has base of area $(3/2)^2\pi$ and height 2. It would be beneficial for you to check answer.

Your nice sketch leads me to
$$V=\frac{1}{2}\int_0^2 \pi [\frac{3- \frac{3}{2}x}{2}]^2 dx$$

I also get the same result but the solution is:
$$V=\int_{0}^{2} \frac{1}{2} \left(\pi \left(\frac{1}{2}-\frac{x}{4}\right)^2\right)dx$$

That's why I am confused at which part I got it wrong, but I think maybe the solution is not correct

Thank you very much anuttarasammyak

 

[anuttarasammyak]

The answer you referred seems to set vertexes are (0,0),(2,0), and (0,1), not (0,3).