0
$\begingroup$

How would I go about the following?

Find the volume of the solid $W$ whose base is the region enclosed by $y=x^2$ and $y=1$, and the cross-sections perpendicular to the y-axis are squares.

$\endgroup$
  • $\begingroup$ hint: $z=2|x|$ and $V=\int \int \int dx dy dz$ $\endgroup$ – Mehrdad Zandigohar Feb 16 '18 at 8:42
0
$\begingroup$

At any position $x$ you have a square of side $1-x^2$. Therefore, you can sum the areas of the squares along $x$. The figure below may help. Thus,

$$V=\int_{-1}^1 (1-x^2)^2~dx=\frac{16}{15}$$

enter image description here

$\endgroup$
  • $\begingroup$ Your sections are perpendicular to $X-$axis. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 19 '18 at 8:16
  • $\begingroup$ @Martín-BlasPérezPinilla Thanks for pointing out my error. I'll just leave this as is, for what it's worth. It does demonstrate a viable method. $\endgroup$ – Cye Waldman Feb 19 '18 at 15:51
0
$\begingroup$

For each $y$ fixed $x$ varies from $-\sqrt y$ to $\sqrt y$, so the integral is: $$ \int_0^1\int_{-\sqrt y}^{\sqrt y}\int_0^{2\sqrt y}\,dz\,dx\,dy = \cdots $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.