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Problem 74 Hard Difficulty

Volume of a solid Consider the region $R$ bounded by the graph of $f(x)=\frac{1}{x+2}$ and the $x$ -axis on the interval $[0,3] .$ Find the volume of the solid formed when $R$ is revolved about the $y$ -axis.

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Video Transcript

this problem. We want to look at a solid formed by rotating the function f of X equal to one over X plus two about the Y axis on the domain where X is and 0 to 3. So what does this look like? Spectrograph. Here's three. If we plug zero end after vex, we get 1/2. And if we plug in three, we get one fist, so it's gonna look something like this decreasing function right there. So I'm gonna rotate it about our y axis best drawing you've ever seen. And we're gonna end up with this, like circus tent shape. So now we know kind of what it looks like, You know, kind of what we have to dio. So we're gonna have to integrate with through the variable. Why, Starting at zero on working our way Don't want half. And we wanna integrate the area of each slice as we go along. So and if you notice, we're gonna have to domains for that, we're gonna have 0 to 1/5 where the radius is constant at three. And when we go past 1/5 the radius is gonna be decreasing all the way to zero. So we can do that is we can find the inverse function of FX, which will be g of Why so if we invert the left and right hand sides will have one over why equal to X plus two. So means will have g of why be won over y minus two where g a fight was X. So we have that function and we can set up our in a role. So first we integrate from 0 to 1/5 for hi are square artistry so high three squared de y and then we're gonna integrate from 1/5. So 1/2 of high r squared where r is g a y, which was won over. Why minus two you are those first new girl just becomes nine pie y from 0 to 1/5 which is just 9/5 pie. Stanford, second in a row should be a squared here inspires square. So we will bring the pile front and will carry out. This will foil this up so 1/5 to 1/2 of and one otherwise just wide of the native one. So it's a little easier. What's will have wide in the negative too minus four. Why did the native one plus for the Y? Now we can use. So the first term is power rule. Ln then power rule again. So bless pie. What's open brackets? Negative. Why? So the negative one minus four. Ln Why? Why is always positive. So we don't need the absolute value. Plus four. Why? And that's on 1/5 to 1/2. Okay, so we can we can evaluate the limits. I will just ignore this ninth spine front for now, So open the bracket, so we'll have negative two minus four Ln of 1/2 plus shoe. Now we'll look at the lower limit. So minus will have negative five minus four Ln of 1/5 plus four vests. Okay, so now we can bring this as we combine like terms. So there's two negative to cancel and distribute. This negative set of possible has plus and minus. We'll bring up high out in front because all the term share pie. So we have nine fits plus five minus 4/5 than for Ellen. Terms we have for Elena to I brought the negative up as a power plus for Ln of 1/5 all right, We'll combine these common denominator five. So we have nine minus four plus 25/5 in the pine front. And for the log we have four. Ellen of two fits. So this becomes 30/5, which is six. So we'll have six pi plus for high Ln of two fits.