Volumes with Cross Sections Practice

Sample questions

5 of 13 — sign in to practice the rest with adaptive difficulty and mastery tracking.

1. Sample 1difficulty 3/5

   Volume of solid $V = \int_a^b A(x)\,dx$ where $A(x)$ is

   • A

     Length of cross-section

   • B

     $f(x)^2$

   • C

     Always $\pi r^2$

   • D

     Cross-sectional area at $x$

     check_circle

   Why

   General cross-section formula for volume.

   View full solution & statsarrow_forward

2. Sample 2difficulty 3/5

   Base on $[0, 1]$ has width $f(x) = 2x$ at each $x$. Semicircular cross-sections perpendicular to $x$-axis (with diameter the base width). Volume?

   • A

     $\tfrac{2\pi}{3}$

   • B

     $\tfrac{\pi}{3}$

   • C

     $\tfrac{\pi}{6}$

     check_circle
   • D

     $\pi$

   Why

   Radius $r = f(x)/2 = x$. Area = $\tfrac{1}{2}\pi x^2$. $V = \int_0^1 \tfrac{\pi x^2}{2}\,dx = \pi/6$.

   View full solution & statsarrow_forward

3. Sample 3difficulty 3/5

   For semicircular cross-sections perpendicular to the $x$-axis with diameter equal to the base width $w(x)$, the area of one cross-section is

   • A

     $\pi w(x)$

   • B

     $\dfrac{\pi w(x)^2}{2}$

   • C

     $\dfrac{\pi w(x)^2}{8}$

     check_circle
   • D

     $\pi w(x)^2$

   Why

   Radius = $w/2$. Semicircle area = $\tfrac{1}{2}\pi r^2 = \tfrac{\pi w^2}{8}$.

   View full solution & statsarrow_forward

4. Sample 4difficulty 3/5

   Base bounded by $y = 1 - x^2$ and $y = 0$. Cross-sections perpendicular to the $x$-axis are squares with side equal to the height of the base ($1 - x^2$). Volume:

   • A

     $\tfrac{16}{15}$

     check_circle
   • B

     $\tfrac{4}{3}$

   • C

     $1$

   • D

     $\tfrac{2}{3}$

   Why

   Side $= 1 - x^2$. Area $= (1 - x^2)^2$. $V = \int_{-1}^{1}(1 - x^2)^2\,dx = 2\int_0^1(1 - 2x^2 + x^4)\,dx = 2(1 - 2/3 + 1/5) = 2(8/15) = 16/15$.

   View full solution & statsarrow_forward

5. Sample 5difficulty 3/5

   Base on $[0, 2]$ has height $1$. Cross-sections are isoceles right triangles with leg = base width = $1$. Volume:

   • A

     $2$

   • B

     $\tfrac{1}{2}$

   • C

     $\tfrac{1}{4}$

   • D

     $1$

     check_circle

   Why

   Triangle area $= \tfrac{1}{2}(1)(1) = 1/2$. $V = \int_0^2 \tfrac{1}{2}\,dx = 1$.

   View full solution & statsarrow_forward