Let R be the region enclosed by the graphs of $y = {(64 x)}^{\frac{1}{4}}$ $y=(64x)^(1/4)$ and $y = x$ $y=x$ . How do you find the volume of the solid generated when region R is revolved about the x-axis?

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Let R be the region enclosed by the graphs of $y = {(64 x)}^{\frac{1}{4}}$ $y=(64x)^(1/4)$ and $y = x$ $y=x$ . How do you find the volume of the solid generated when region R is revolved about the x-axis?

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Answer 1 · 2017-06-30T19:40:23

$\frac{64 π}{3}$ $(64pi)/3$

Explanation:

First, plot the two graphs, shading the bounded region between them.

![Desmos.com]( useruploads.socratic.org )

Second, recall the formula for solid of revolution about the $x$ $x$ -axis

$\int_{x_{0}}^{x_{f}} π {(outer radius)}^{2} - π (inner radius) 2 d x$ $int_(x_0)^(x_f)pi("outer radius")^2-pi("inner radius")2dx$

The outer radius is the height from the $x$ $x$ -axis to the outer most graph, or $y = {(64 x)}^{\frac{1}{4}}$ $y=(64x)^(1/4)$

$\int_{x_{0}}^{x_{f}} π {({(64 x)}^{\frac{1}{4}})}^{2} - π {(inner radius)}^{2} d x$ $int_(x_0)^(x_f)pi((64x)^(1/4))^2-pi("inner radius")^2dx$

The inner radius is the height from the $x$ $x$ -axis to the inner most graph, or $y = x$ $y=x$

$\int_{x_{0}}^{x_{f}} π {({(64 x)}^{\frac{1}{4}})}^{2} - π {(x)}^{2} d x$ $int_(x_0)^(x_f)pi((64x)^(1/4))^2-pi(x)^2dx$

The lower limit of integration is the smallest $x$ $x$ -value where the two curves meet, or $x_{0} = 0$ $x_0=0$ . The upper limit of integration is the largest $x$ $x$ -value where the two curves meet, or $x_{f} = 4$ $x_f=4$

$\int_{0}^{4} π {({(64 x)}^{\frac{1}{4}})}^{2} - π {(x)}^{2} d x$ $int_(0)^(4)pi((64x)^(1/4))^2-pi(x)^2dx$

$\int_{0}^{4} π {(64 x)}^{\frac{1}{2}} - π x^{2} d x$ $int_(0)^(4)pi(64x)^(1/2)-pix^2dx$

$\int_{0}^{4} π 8 x^{\frac{1}{2}} - π x^{2} d x$ $int_(0)^(4)pi8x^(1/2)-pix^2dx$

$\int_{0}^{4} π 8 x^{\frac{1}{2}} d x - \int_{0}^{4} π x^{2} d x$ $int_(0)^(4)pi8x^(1/2)dx-int_(0)^(4)pix^2dx$

$8 π \int_{0}^{4} x^{\frac{1}{2}} d x - π \int_{0}^{4} x^{2} d x$ $8piint_(0)^(4)x^(1/2)dx-piint_(0)^(4)x^2dx$

$\frac{16}{3} π {[x^{\frac{3}{2}}]}_{0}^{\frac{3}{2}} - \frac{π}{3} {[x^{3}]}_{0}^{3}$ $16/3pi[x^(3/2)]_(0)^(4)-pi/3[x^3]_(0)^(4)$

$\frac{16}{3} π {(4)}^{\frac{3}{2}} - \frac{π}{3} {(4)}^{3} = \frac{64 π}{3}$ $16/3pi(4)^(3/2)-pi/3(4)^3=(64pi)/3$

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Let R be the region enclosed by the graphs of $y = {(64 x)}^{\frac{1}{4}}$ $y=(64x)^(1/4)$ and $y = x$ $y=x$ . How do you find the volume of the solid generated when region R is revolved about the x-axis?

1 Answer

Explanation:

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Let R be the region enclosed by the graphs of y=(64x)14y=(64x)^(1/4) and y=xy=x. How do you find the volume of the solid generated when region R is revolved about the x-axis?

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Let R be the region enclosed by the graphs of $y = {(64 x)}^{\frac{1}{4}}$ $y=(64x)^(1/4)$ and $y = x$ $y=x$ . How do you find the volume of the solid generated when region R is revolved about the x-axis?