Question

The base of a solid is a circle x^2+y^2=r^2. Find the volume of the solid given that the cross sections perpendicular to the x-axis are squares

Answer 1

It says the answer is (16/3)r^3 I just do not get how to put this in an integral

Answer 2

I'm not sure I understand what solid the problem is describing.

Answer 3

Well all I know is that cross sections are like parts of the shapes so it should look like this i guess

Answer 4

Im not to sure if Im correct

Answer 5

The circle is the base and the squares I guess are thin sections of it

Answer 6

Is that the entire question you posted in the original post?

Answer 7

yes except part b) asks for equalateral triangles but I only have to do part a)

Answer 8

I found this if it helps

Answer 9

OK, so the general form of the integral is:\[\int\limits_a^b A(x)\ dx\]We're integrating across x, and the min and max values for x are -r to r, so that's what out limits will be.

We also need to find a function for the area as a function of x. At any given x, y ranges from the bottom of the circle to the top of the circle. To get that distance, we take the top semi-circle minus the bottom semi-circle:\[\sqrt{r^2-x^2}-(-\sqrt{r^2-x^2})=2\sqrt{r^2-x^2}\]Does that much make sense?

Answer 10

yes

Answer 11

OK, this distance represents the base of the square. And to find the area of a square, you just take its base squared:\[A(x)=(2\sqrt{r^2-x^2})^2=4\sqrt{r^2-x^2}\]Go back to our integral:\[V=\int\limits_{-r}^{r}4\sqrt{r^2-x^2}\ dx\]

Answer 12

Sorry.

Answer 13

It should be:\[V=\int\limits_{-r}^{r}4(r^2-x^2)\ dx\]

Answer 14

Ok I getcha

Answer 15

And integrating that, you'll get (16/3)r^3

Answer 16

Thanks dude for your time!!! :D I got it right