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The region enclosed by the given curves is rotated about the specified line. Find the volume of the resulting solid.$$y=x^{3}, y=\sqrt{x} ; \quad \text { about } y=1$$

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$=\frac{10 \pi}{21}$

Calculus 2 / BC

Chapter 7

APPLICATIONS OF INTEGRATION

Section 2

Volumes

Applications of Integration

Marielis M.

January 31, 2022

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 5x, y = 5 x ; about y = 5

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = 5x, y = 5 Sqrt(x) ; about y = 5

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The region enclosed by the…

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The region bounded by the …

Find the volume of the sol…

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I want to find the volume bounded by these two equations. So, first of all, I need to think about what the area is between these two equations and that area happens here. Next. I need to know what is the access of revolution in this case, we're gonna rotate around the line y equals one so we can take this whole two dimensional area and then spin it up around the line. White was one so revolving in three dimensions you'd get some sort of open figure. You're going to see, like this cut out in the volume from that white space that we don't have filled in from right here because we have that missing space between the shaded area and the access of revolution. This is a washer set up, so because it's the washer set up, I'm gonna go and write that out. It's pi times the integral R squared minus little r squared. And then I want to use whatever variable is the axis here, and I actually use d X because Y equals one is like the x axis. Notice how those air parallel. So we use X as the very well for integration which means we also use X on our bounds and in that capital R and lower case R. So now we need to decide What are those X values? Well, the area is bounded by these graphs of X, cubed and X and sorry, the square root of X. And you may know that those graphs share points at Here's your own 11 If you weren't able to just inspect that and know that you could solve this equation X cubed equals the square root of accidents. See where those values have the Sorry where those equations have the same value. So my X one happens at Deer O and X two happens at one. So over here on my Anna, bro, I'm gonna go from 0 to 1 next. I need to know Capital R and lower case R Capital are is what if nothing was missing? Well, nothing would be missing if this volume was all the way filled in from the axis to the edge of the figure. So that would mean going from y equals one down to the curved graph and the curved draft. X cubed is the bottom of that figure. See how I have it all filled in. I always think about that like Upper minus lower when I'm going in this direction. So I started with one first down to the execute. So that's my capital art. And then for lower case R is what are we taking away? Well, the stuff that we're taking away here is this portion. That's the empty space between the axis and the shaded area. So upper minus lower again that goes from one down to the square root graph. If you follow that line, that goes to the square root graph so one minus the square root graph is that lower case R and now I have my integral totally set up. This is a good time if I can to type the whole thing into calculator or online system. If not, I would need to expand thes by no meals, squaring each term and member of you square these terms, it means you're multiplying at times itself, so you have to foil or distribute to get all those terms expanded, which is gonna take some time. So for the sake of time here, I'm going to calculate this expression one minus X cubed squared minus one minus square root of X squared going from 0 to 1. And the whole thing is actually 10/21 I still have the pie in front. So for my final answer, I'd write 10 pi over 21 and that's it.

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