Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

enter image description here

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.

share|improve this question
    
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

1 Answer 1

up vote 0 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.