Graph function

[Steven G]

that's the exact problem. if you mean book copy, i do not have it unfortunately.
as for 3, as far as i know you calculate the volume by Pi * integral of (function of the curve)^2, where integral is defined from the start point on x axis to the end point of the x axis of the curve. I go from 0 to 3 because i don't think they asked for infinity answer and because it's overall a different shape.

What is overall a different shape?? You can integrate or calculate volume for any shape (volume if you have symmetry). That is just what integration is for--to calculate areas, volumes, surface areas of many non-standard curves.

Look, if the problem said to find the volume of the enclosed area that is one problem, if it said to find the volume that is another problem. You need to decide which one you want to do. Before you do that, no one can help you.

 

[BigBeachBanana]

that's the exact problem. if you mean book copy, i do not have it unfortunately.
as for 3, as far as i know you calculate the volume by Pi * integral of (function of the curve)^2, where integral is defined from the start point on x axis to the end point of the x axis of the curve. I go from 0 to 3 because i don't think they asked for infinity answer and because it's overall a different shape.

What is the given answer?

 

[Loki123]

What is overall a different shape?? You can integrate or calculate volume for any shape (volume if you have symmetry). That is just what integration is for--to calculate areas, volumes, surface areas of many non-standard curves.

Look, if the problem said to find the volume of the enclosed area that is one problem, if it said to find the volume that is another problem. You need to decide which one you want to do. Before you do that, no one can help you.

I don't know. It just said volume.

 

[Loki123]

What is the given answer?

3Pi/4

 

[BigBeachBanana]

3Pi/4

This is the graph of [imath]y=-\frac{\sqrt{x}}{3}\cdot\left(x-3\right),\, x \in [0,3][/imath]


Rotating around the x-axis:

The Volume:
[math]V=\pi\int_{0}^{3}\left(-\frac{\sqrt{x}}{3}\cdot\left(x-3\right)\right)^2\,dx=\pi\int_{0}^{3} \frac{x\cdot(x-3)^2}{9}\,dx =\frac{3\pi}{4}[/math]

 

[Loki123]

This is the graph of [imath]y=-\frac{\sqrt{x}}{3}\cdot\left(x-3\right),\, x \in [0,3][/imath]
View attachment 32386

Rotating around the x-axis:
View attachment 32387
The Volume:
[math]V=\pi\int_{0}^{3}\left(-\frac{\sqrt{x}}{3}\cdot\left(x-3\right)\right)^2\,dx=\pi\int_{0}^{3} \frac{x\cdot(x-3)^2}{9}\,dx =\frac{3\pi}{4}[/math]

Thank you! I get it now!