Calculus 2 Volume

[Tulay]

Can someone please help me solve this? First using washer method then the shell method?
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[tkhunny]

Have you tried anything? Show us what you have and we can help guide you on your way..

 

[Dr.Peterson]

Can someone please help me solve this? First using washer method then the shell method?
-View attachment 25636

Please show us what you've tried, so we can see what sort of help you need.

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The hardest part is usually to keep track of where the axis is, in order to get the right radii. They will be the difference between x-coordinates on the appropriate part of each curve and on the axis.

I don't think I'd prefer the washer method for this one!

 

[Tulay]

Have you tried anything? Show us what you have and we can help guide you on your way..

 

[Dr.Peterson]

View attachment 25637

You solved both equations for x, but forgot the $\pm$. And in your integral, you didn't make r₁ and r₂ functions of y, and just used the maximum value of each.

You'll need two separate integrals, one from y=1 to 2, and another from y=2 to 3. Each will take r₁ at the right side and r₂ at the left side of the same curve.

 

[HallsofIvy]

Notice that the "washer method" is really two "disk" methods:
$\displaystyle \pi \int_a^b (f^2- g^2)x= \pi\int_a^b f^2dx- \pi\int_a^b g^2dx$.

First find the volume when $\displaystyle y= 3- x^2$, for x between -1 and 1, rotated around y= -1, $\displaystyle \pi\int_{-1}^1 (3- x^2- (-1))^2 dx= \pi\int_{-1}^1 (4- x^2)^2 dx$.

Then find the volume when $\displaystyle y= x^2+ 2$, for x between -1 and 1, rotated about y= -1, $\displaystyle \pi\int_{-1}^1 (x^2+ 2- (-1))^2 dx= \pi\int_{-1}^1 (x^2+ 3)^2 dx$.

And subtract!

 

[Tulay]

You solved both equations for x, but forgot the $\pm$. And in your integral, you didn't make r₁ and r₂ functions of y, and just used the maximum value of each.

You'll need two separate integrals, one from y=1 to 2, and another from y=2 to 3. Each will take r₁ at the right side and r₂ at the left side of the same curve.

You solved both equations for x, but forgot the $\pm$. And in your integral, you didn't make r₁ and r₂ functions of y, and just used the maximum value of each.

You'll need two separate integrals, one from y=1 to 2, and another from y=2 to 3. Each will take r₁ at the right side and r₂ at the left side of the same curve.

Like this?

 

[Dr.Peterson]

Like this?

I don't see that you've done what I said. Can you explain each part of this?



What are your two radii, and where do they show up on the graph?