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Find the volume of the solid whose base is the region in the first quadrant bounded by $y=x^2$, $y=1$, and the $y$-axis and whose cross-sections perpendicular to the $x$ axis are squares. Also for a semicircle.

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Squares: The area of a square is $s^2$ where $s$ is a side of the square. Therefore, each cross-section perpendicular to the $x-axis$ is going to have the area $(1-2x)^2$ since those are the two lines bounding the region. To find the volume of the solid, we have to find the $x$ values of the intersections of all the lines bounding the region. $$2x=1, \space 2x=0$$ $$x=\frac{1}{2}, \space x=0$$ These $x$ values are the bounds of the integral we're going to take to find the volume, which is $$\int_{0}^{\frac{1}{2}}(1-2x)^2dx$$ We do a $u$ substitution to have $u=1-2x$. $$\frac{du}{-2}=dx$$ We substitute the original integral to receive $$\frac{1}{2}\int_{0}^{1}u^2du=\frac{1}{6}$$This is the volume when the cross-sections are squares.

Semi-circles: The area of a semi-circle is $\frac{\pi r^2}{2}$ where $r$ is the radius of the circle. Using our example above, $r=\frac{1-2x}{2}$, and we can simply find the volume again using the integral $$\pi\int_{0}^{\frac{1}{2}}(\frac{1-2x}{2})^2dx$$ which is equal to $\frac{\pi}{24}$. Hope this helps.

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I doubt the phrasing of the question is the reason for the downvotes. Copy and paste of homework questions with no effort shown is generally frowned upon by the community. –  Austin Mohr Jan 14 '13 at 5:32
Oh, okay. Kind of new to the community. Thanks for the info –  joejacobz Jan 14 '13 at 5:52
No effort is hardly the case. I have spent nearly seven hours on this problem. Thanks for the help! It still says my answer is wrong :/ –  user58204 Jan 14 '13 at 11:49
Welcome to Math.SE user58204, You can accept the answer of @joejacobz like the best, by clicking the green accepted symbol. This, to recognize the effort he made ​​to answer your question. –  dwarandae Jan 14 '13 at 16:23
Shouldn't the boundary of y=x^2 figure into your solution? Or have I completely misunderstood the geometry of the problem –  User58220 Jan 14 '13 at 20:36
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