Find the volume $y=x^2 [0,2]$

Question

Find the volume of the solid whose base is the region bounded between the curve $y=x^2$ and $x$-axis $[0,2]$ and whose cross-section taken perpendicular to the $x$-axis are squares.

My solution

$$V = \int_0^2\pi f(x)^2\,\mathrm dx = \int_0^2\pi x^4\,\mathrm dx = \frac{32}5\pi$$

But in the book, the answer is $\dfrac{32}5$. Why?

You are correct, and apply Fahrenheit 451 to that book if it contains many more mistakes... — DonAntonio, Dec 24 '13 at 16:08
There is no $\pi$; it is not a solid of revolution. See @Suburban13's answer. — Gonçalo, Jan 05 '24 at 11:49

score 0 · Answer 1 · edited Jan 05 '24 at 12:22

0

"whose cross-section taken perpendicular to the $x$-axis are squares." You'd have to take small squares and integrate them over the region, so a small square would be $(x^2)^2=x^4$ and integrating them over the interval would give the volume they are asking in the question.

edited Jan 05 '24 at 12:22

Gonçalo

9,312

answered Jan 05 '24 at 10:46

Suburban13

23

1

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – José Carlos Santos Jan 05 '24 at 11:07

Find the volume $y=x^2 [0,2]$

1 Answers1