Volume integration using washermethod

[peercortsa]

volume integration using "washer method"

Homework Statement

The region R bounded by y = x$$^{2}$$, y = 0, x = 1 and x = 4 is rotated about x = −1

Homework Equations

i know that these equations take on the form of $$\pi\int_a^b \\((outer radius)^{2} - (inner radius)^{2})\\,dx$$

The Attempt at a Solution

so i set the problem up like this and still can't get the correct answer which is $$339\pi/2$$

-for the bounds i know that the integral i want to evaluate is between 0 and 16 based on the line y=0 and the function $$y=x^{2}$$ evaluated at x=4.

-the outer radius is $$1+\sqrt{y}$$ and the inner radius is just 1

$$\pi\int_0^{16}\\ (1+\sqrt{y})^2-(1)^2\\,dy$$

 

[tiny-tim]

Welcome to PF!

The region R bounded by y = x$$^{2}$$, y = 0, x = 1 and x = 4 is rotated about x = −1
…
i know that these equations take on the form of $$\pi\int_a^b \\((outer radius)^{2} - (inner radius)^{2})\\,dx$$

Hi peercortsa! Welcome to PF!

That's right … but then your thickness is dx, so that's what you must integrate over …

∫(blah blah) dx, where blah blah is entirely a function of x.

 

[peercortsa]

so... did i even set up the equation correctly because i still can't seem to get the right answer no matter what i try

 

[tiny-tim]

so... did i even set up the equation correctly because i still can't seem to get the right answer no matter what i try

Your π∫…dx is correct for the area …

but you still need to put the height inside the ∫ to make the volume.

Have a go!