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Find the volume of the described solid $ S $.

The base of $ S $ is the same base as in Exercise 58, but cross-sections perpendicular to the x-axis are isosceles triangles with height equal to the base.

$\frac{8}{15}$ units $^{3}$

Applications of Integration

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were given a solid S. And we have to find the volume of this solid. The stars. We're told that the base of the solid is the same base has an exercise 58 but the cross sections perpendicular to the X axis are isosceles triangles with a height equal to the base. So from exercise 58 the base was actually a parabola. It's the parabola mm uh Y equals one minus X squared and the line Y equals zero in text you a lot. Okay? He said fucking jews control right? Imagine this is saying it as okay. Yes. So first you want to find the area of one of these triangles, right to do this, we need to find what the bases and to find the base truck called B. This is just the value of why? Which is one minus X. Squared. So the top curve minus the bottom curve we'll keep. Mhm. Now their base. Well, we're told the area of the assassins triangle. This is one half times the base times the height. And we're told that the height is equal to. Don't fuck with me bro. I was. And base this is the same as one half B squared which is one half times one minus x squared squares, racist. So this is the area of one of the triangles. In terms of X. Using the cross section interpretation, our volume is the integral From what we see on our base. The X range is from negative 12 positive one of our area A. Of X. Dx. This is one half times the integral from negative one positive one of one minus X squared squared ultimate dx. Yeah. Now if you foil this out and take anti derivatives, this is a pretty easy integral to evaluate. And you should get 8/15. Did you could get sucked wrong. Mhm.

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Applications of Integration