Find the volume for a solid whose base is the region between the curve y = sin x y=\sqrt{\sin x} and the x-axis on the interval from [ 0 , π ] \left\lbrack0,\pi\right\rbrack and whose cross sections are equilateral triangles with bases parallel to the y-axis.

this problem, we're asked to find the volume for a solid whose base is the region between the curve y equals square root of sine of x and the x axis on the interval from zero to pi and whose cross sections are equilateral triangles with bases parallel to the y axis. Now there seems to be a lot going on in this problem, so let's go ahead and break this down and start by sketching our solid. So I'm going to go ahead and draw my tilted axes here. And then we know we're looking at the interval from zero to pi. So I know I have my origin point zero and I'm going to go ahead and mark pi on my x axis there. Now, if we graph the square root of sine of x from zero to pi, it forms this sort of semicircle shape. So with this semicircle, we're going to go ahead and extend it upwards to form our solid. So with that solid formed here, I'm going to go ahead and draw that semicircle again. We have that solid with that base formed by the curve y equals square root of sine of x and the x axis from zero to pi. Now the second part of this problem tells us that the cross sections as we slice into this solid parallel to the y axis, these cross sections are equilateral triangles. So I'm going to go ahead and turn this triangle facing us to the front so that we can really see what's going on here. So with this equilateral triangle, equilateral just means that all of our sides are equal to each other. So how exactly with all of this knowledge and our sketch here are we going to set up our volume integral? We know that we need to integrate the area function of this equilateral triangle cross section. And we know that the area of a triangle is equal to one half base times height. So we need to figure out what the base and height of our triangle actually is. Now coming back over to our graph here, this semicircle shape is our function, the square root of sine of X. Now we're gonna refer to this function here as a Y as it was given to us in the problem. So here with our function, we can see that the base of our triangle goes from our X axis to our function. So that makes our base equal to that function y. Now because this is an equilateral triangle, that means that all of my side lengths are y. So I know that my base is y. So as I fill in my area function, this is equal to one half times y times something. And we need to figure out what that height, what that something is. So coming back over here to our triangle, if I go ahead and draw a dotted line here for my height, this splits my equilateral triangle into two right triangles. And with this right triangle, we have a base of one half y because we're splitting this triangle in half. So that means that we can then figure out what this height is using the Pythagorean theorem. So I'm going to go ahead and set up my Pythagorean theorem here. Remember, a squared plus b squared equals c squared. So my two side lengths here, I have one half y squared plus h squared equals Y squared. So what I've done here is I've taken A squared plus B squared, set that equal to C squared. So now what are we actually solving for here? Well, we want to figure out what H is because we already know what Y is. So let's go ahead and solve for h here. I'm going to start by subtracting over this one half y squared to cancel it out and leave me here with h squared equals y squared minus one half y squared. Now I'm going to go ahead and distribute that squared into my parenthesis here. So this gives me y squared minus one fourth y squared, going ahead and squaring that one half. So if h squared is equal to this, I can go ahead and simplify this further. If I take y squared and subtract one fourth y squared, that ultimately gives me three fourths y squared and that's equal to h squared. Now the only thing we have left to do here to get this height by itself is to square root both sides. So that gives me that h is equal to the square root of three fourths y squared, which we can further simplify here to the square root of three over two y because that radical will cancel out that squared. And when we take the square root of three over four, that is root three over two. So now that we know what our height is, we can put that into our area function. So this is one half Y times root three over two Y. Now we can go ahead and multiply all the way across here to ultimately give us root three over four y squared. Now we can go ahead and plug in what y actually is because we know that it's the square root of the sine of x. So this gives us root three over four times root sine of x squared and that squared nicely cancels that square root. So this is root three over four sine of x and this gives us our area function. Now really this was the most complicated part of this problem because now we can just stick this area function that we spent all this time finding into our volume integral. So let's go ahead and set up our volume integral here. I'm going to go ahead and get out of the way. So our volume here is going to be equal to the integral from zero to pi. Remember those bounds were given to us in our problem. Zero to pi of that area function root three over four sine x d x. Now we can actually evaluate that integral here. So I have a root three over four times negative cosine of x. So I'm going to stick that negative on the outside here Bounded from zero to Pi. Now just plugging those bounds in here, we get negative root three over four times the cosine of Pi minus the cosine of zero. Now the cosine of Pi is negative one. The cosine of zero is is positive one. So we have negative one minus one that is negative two. So we get here negative root three over four times negative two, which we can ultimately simplify to be a positive root three over two for our final answer here. Now this isn't just a random number. Remember this actually represents the volume of this solid that has these equilateral triangle cross sections. Now Now if you have questions about how I set up this integral or did any of this algebra along the way, feel free to let us know in the comments. Thanks for watching. I'll see you in the next one.

Find the volume for a solid whose base is the region between the curve y = sin ⁡ x y=\sqrt{\sin x} and the x-axis on the interval from [ 0 , π ] \left\lbrack0,\pi\right\rbrack and whose cross sections are equilateral triangles with bases parallel to the y-axis.

− 4 + 3 4 -\frac{4+\sqrt3}{4}

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

Calculus

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

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Multiple Choice

Find the volume for a solid whose base is the region between the curve $y = \sqrt{\sin x}$ and the x-axis on the interval from $open bracket 0 comma pi close bracket$ and whose cross sections are equilateral triangles with bases parallel to the y-axis.

$\frac{\sqrt{3}}{2}$

$- \frac{\sqrt{3}}{4}$

$- \frac{\sqrt{3}}{2}$

$- \frac{4 + \sqrt{3}}{4}$

Verified step by step guidance

First, identify the region of integration. The base of the solid is the region between the curve y = \sqrt{\sin x} and the x-axis on the interval [0, \pi].

Next, determine the shape of the cross sections. The problem states that the cross sections perpendicular to the x-axis are equilateral triangles with bases parallel to the y-axis.

Calculate the side length of the equilateral triangle at a given x. The side length is equal to the value of the function y = \sqrt{\sin x} at that x.

Find the area of an equilateral triangle with side length s. The formula for the area of an equilateral triangle is \frac{\sqrt{3}}{4} s^2.

Set up the integral to find the volume. Integrate the area function A(x) = \frac{\sqrt{3}}{4} (\sqrt{\sin x})^2 from x = 0 to x = \pi to find the total volume.