Find the volume of the described solid

[lalashi485]

The solid lies between planes perpendicular to the x-axis at x=0 and x=7. The cross sections perpendicular to the x-axis between these planes are squares whose bases run from the parabola y=-2√x to the parabola y=2√x.

I need some help in case I'm doing something wrong. This is from a review that I am doing and I keep getting the answer as 49 but in the back it says it is suppose to be 392. I apologize if I didn't post this correctly. This is my first time on this forum

To start I sketched the graph (which I wish I could do here). Then I worked out the problem as follows:

2√x - (-2√x) = 4√x

A(x)= (4√x)^{2
=4x
=4x/2
=2x}

V=₀∫⁷ 2x dx

2x²/2 = x²

x²| from 0 t0 7

[(7)²]-[(0)²]

V=49

 

[wjm11]

The solid lies between planes perpendicular to the x-axis at x=0 and x=7. The cross sections perpendicular to the x-axis between these planes are squares whose bases run from the parabola y=-2√x to the parabola y=2√x.

I need some help in case I'm doing something wrong. This is from a review that I am doing and I keep getting the answer as 49 but in the back it says it is suppose to be 392. I apologize if I didn't post this correctly. This is my first time on this forum

To start I sketched the graph (which I wish I could do here). Then I worked out the problem as follows:

2√x - (-2√x) = 4√x

A(x)= (4√x)^{2
=4x
=4x/2
=2x}

V=₀∫⁷ 2x dx

2x²/2 = x²

x²| from 0 t0 7

[(7)²]-[(0)²]

V=49

Several comments: You are not trying to find the area between to functions, and your algebra is in error anyway; (4√x)^2 does not equal 2x, but that is beside the point.

Secondly, you correctly described you shape as a parabola. In light of that, consider your integral: ₀∫⁷ 2x dx. Do you see that this integral is finding the area under the line y = 2x?
***
You are trying to find a volume of revolution. You should look at examples of the "disc" or "washer" method in your book. Just use the function y=2√x and rotate it about the x-axis, from 0 to 7.

 

[HallsofIvy]

This NOT a volume of revolution. You are told that cross sections are "squares whose bases run from the parabola y=-2√x to the parabola y=2√x." Okay, such cross sections are squares with side length \(\displaystyle 2\sqrt{x}- (-2\sqrt{x})= 4\sqrt{x}\) and so area \(\displaystyle (4\sqrt{x})^2= 16x\). If you imagine the solid divided into many such square slabs (as in a Riemann sum) each has thickness "\(\displaystyle \Delta x\)" and so volume \(\displaystyle 16x \Delta x\). The volume of the entire figure is approximated by the sum of such things, \(\displaystyle \sum 16x \Delta x\). In the limit, that Riemann sum becomes the integral \(\displaystyle \int_0^7 16x dx\).

 

[wjm11]

This NOT a volume of revolution. You are told that cross sections are "squares whose bases run from the parabola y=-2√x to the parabola y=2√x." Okay, such cross sections are squares with side length \(\displaystyle 2\sqrt{x}- (-2\sqrt{x})= 4\sqrt{x}\) and so area \(\displaystyle (4\sqrt{x})^2= 16x\). If you imagine the solid divided into many such square slabs (as in a Riemann sum) each has thickness "\(\displaystyle \Delta x\)" and so volume \(\displaystyle 16x \Delta x\). The volume of the entire figure is approximated by the sum of such things, \(\displaystyle \sum 16x \Delta x\). In the limit, that Riemann sum becomes the integral \(\displaystyle \int_0^7 16x dx\).

Oops. Sorry for the misread; I blanked out the entire "squares part". Thank you, Halls, for catching this.

Oh, well... off to the corner...:-?