Calc Help on volume from known cross section areas (SS will be posted shortly)

Question

darkknight · Answer

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darkknight · Answer

Very confused on what to do here, I originally graphed them out and saw that one was a top function and the other was below it so I wrote the equations in terms of x
so that $y=9-\sqrt{\left(-18+x\right)}$ and $y=3+\sqrt{54-x}$

darkknight · Answer

Volume is the integral of the area $$V=\int\limits_{18}^{54}A(t)*dt$$
And Area of a semi-circle is $$(\pi*r^2)/2$$ but idrk what to do next

darkknight · Answer

@imqwerty @dude

darkknight · Answer

A problem sort of similar to this, I used the same ideas as I did above
$$V=\int\limits_{-.347}^{1}\left(\left(e^{2.2x}-\left(2.2x^{2}+.2\right)\right)^{2}\right)dx$$
because on the left both functions intersect at x=-.347 and we are bounded by x=1
and since cross sections are squares the area is length squared hence why I found the distance b/w the 2 functions and then squared it for area. I just don't get what radius is in the other problem is it like half the distance? Not sure how to set the other one up

darkknight · Answer

@563blackghost e.e

darkknight · Answer

@florisalreadytaken @AZ @vocaloid

Vocaloid · Answer

Looking @ my notes there's a specific formula for the volume of a shape w/ semicircular cross sections perp. to x

$$V = \frac{ \pi }{ 8 }\int\limits_{a}^{b}(t-b)^{2}dx$$

obv. change dx to dy since ours are perpendicular to the y-axis. additionally, your a and b boundary values are also the y-values of the boundaries (which are 3 and 9, not 18 and 54. Got these by graphing the boundary equations.)

When I plugged in everything into the formula I was able to obtain the answer key solution.

darkknight · Answer

Thanks for the help! Just some clarifying questions, 
1. so if we are finding cross-sections perpendicular to say the y-axis we use dy instead of dx and is finding cross-sections perpendicular to y-axis same as parallel to x-axis?
2. How does this actually look like? like is it semi-circles coming out of the page or...?
3.

darkknight · Answer

Hmm, I now realized t-b is twice the radius. t-b (which equals $$-2y^2+24y+-54$$)
So the radius is $$(-2y^2+24y+-54)/2$$) now we can put this in the semi-circle formula $$\pi*r^2/2$$ And I got the same answer the formula gave me, and I can use this process on other shapes such as triangular cross-sections, etc

imqwerty · Answer

@darkknight wrote:

Thanks for the help! Just some clarifying questions, 1. so if we are finding cross-sections perpendicular to say the y-axis we use dy instead of dx and is finding cross-sections perpendicular to y-axis same as parallel to x-axis? 2. How does this actually look like? like is it semi-circles coming out of the page or...? 3.

1) Yes. In general, we pick a direction in which we can break down the given shape into smaller fundamental parts (like semicircular slices in this question) and do the integration in that direction. If we're in a 2D plane, then yes, perpendicular to y is parallel to x. However, if we're in a 3D, this may not necessarily hold. For example, the z-axis is perpendicular to both the x and y axes. 2) Yes, semicircles coming out / going into the page.

darkknight · Answer

Thanks

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