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I have to find the Area of the Vertical cross section A and the Volume. I have no idea how to do this problem we never learned this in class. Need all the help I can get. Thank you.

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Text is preferred over images when posting questions. Images are impossible to find via search. You help others who are looking for the same answer by posting the text instead. – Ayman Hourieh Feb 12 '13 at 0:43
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Suppose that the height $h$ and the base $b$ of a triangle are equal. Since the area of the triangle with base $b$ and height $h$ is $(1/2)bh$, the area of a base equals height triangle is $(1/2)b^2$, where $b$ is the base.

When $x=2$, we have $y^2=25-2^2=21$. So the base "at" $x=2$ is $2\sqrt{21}$, and therefore the area of cross-section is $(1/2)(2^2)(21)$.

In general, the base "at" $x$ of our triangle is $2\sqrt{25-x^2}$. Thus the area $A(x)$ of cross section at $x$ is $(1/2)(2\sqrt{25-x^2})^2$, which simplifies to $2(25-x^2)$.

To find the volume, you will need to integrate $A(x)\,dx$ from the beginning ($x=-5$) to the end ($x=5)$. It may make life easier to exploit symmetry, integrate from $0$ to $5$, and double the result.

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Thank you that helped a lot. Any idea how I would find the volume? – Ak47 Feb 12 '13 at 0:45
It is almost explicitly in the post. The volume is $\int_{-5}^5 2(25-x^2)\,dx$. The integration is easy. I suggested in the post using instead $2\int_0^5 2(25-x^2)\,dx$. – André Nicolas Feb 12 '13 at 0:48

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