Sections of this figure, that our vertical, I should say our perpendicular to the X axis, those cross sections are going to be isosceles right triangles. Sections of the figure, that's what this yellow line is. What I've drawn here in blue, you could view this kind of the top ridge of the figure. Lets see if we can imagine a three-dimensional shape whose base could be viewed as this shaded in region between the graphs of Y is equal to F of X and Y is equal G of X. ![]() Your bounds should obviously be the least and greatest x-values that lie on the circle. You should have the base length from the previous step, which is all you need to find the cross-sectional area.Ĥ. The cross-section is an equilateral triangle, and you probably learned how to calculate the area for one of those long ago. Remember that to express a circle in terms of a single variable, you need two functions (one for above the x-axis and one for below it, in this case).ģ. A width dx, then, should given you a cross-section with volume, and you can integrate dx and still be able to compute the area for the cross-section. You know the cross-section is perpendicular to the x-axis. Integrate along the axis using the relevant bounds.Ī couple of hints for this particular problem:ġ. ![]() Find an expression for the area of the cross-section in terms of the base and/or the variable of integration.Ĥ. Find an expression in terms of that variable for the width of the base at a given point along the axis.ģ. Figure out which axis (and thus which variable) you'll be using for integration.Ģ. 00:07 That's just a terrible looking circle and that's not any better but if you were to draw out i saw these right triangles there with the base being this side so i just draw that a little bit larger and right here being the right angles then this i like to be base root to this side length would be base divided by root 2 so if you think about the area with this being the base and the height the area of a triangle is equal to 1 half times one base times the height and yeah, well, you're looking at then is root 2 times root 2 is 2 times this other 1 half we 1 4th times that base being squared so i guess all you really have to think about then i guess i should draw this out as well that the equation is x squared plus y squared equals 4 that if you're trying to just solve for the top half, it would just be solving for y so you'd be subtracting x squared to the right side and when you square root both sides would be the plus or minus version so the volume i'm going to move that 1 4th in front the integral will be from negative 2 to 2 i guess i should explain that in a circle that 4 is equal to r squared so the radius of this thing is going from negative 2 to 2 that's how i got my bounds square root of 4 is 2 and so the upper function is the positive version of 4 minus x squared and then the lower function is the negative version of 4 minus x squared and then i have to square that piece in terms of x so if i simplify this just a little bit what i have going on is this turns into a plus because subtracting a negative will be plus so it's 2 square roots of 4 minus x squared being squared, but if you square each of those pieces then just cross that out and put it there 2 squared will cancel out with the 4 in front and the square root will cancel out with this square.I won't give you the answer, but I'll offer a general strategy for questions of that variety:ġ.
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