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Chapter 13: Problem 25

Use a double integral to find the volume of the solid bounded by the graphs of the equations. $$ z=x^{2}, z=0, x=0, x=2, y=0, y=4 $$

Short Answer

Expert verified
The volume of the solid is \(\frac{32}{3}\) cubic units.

Step by step solution

01

Understand the Geometrical Meaning

The equations \(z=x^{2}, z=0, x=0, x=2, y=0, y=4\) define a solid that is essentially a 3D region under the curve \(z=x^{2}\) between \(x=0\) and \(x=2\), and \(y=0\) and \(y=4\). The solid is bounded below by \(z=0\) (the xy-plane).
02

Set Up Double Integral

The volume of a solid bounded by a surface \(z=f(x,y)\) and plane \(z=0\) can be calculated through the double integral \(\int_Y \int_X f(x,y)dxdy\). Now, in this case, \(f(x,y)=x^{2}\). Also, \( 0 \leq x \leq 2\) and \(0 \leq y \leq 4\). So, the double integral becomes \(\int_{0}^{4} \int_{0}^{2} x^{2} dx dy\).
03

Compute Inner Integral

The inner integral concentrates on the variable \(x\). It can be computed by using standard integral rules, leading to \(\int_{0}^{2} x^{2} dx = \left[ \frac{1}{3}x^{3} \right]_{0}^{2} = \frac{8}{3}\). The resulting (simplified) integral is \(\int_{0}^{4} \frac{8}{3} dy \).
04

Compute Outer Integral

Now, let's compute the outer integral which is with respect to the variable \(y\). This results in \(\int_{0}^{4} \frac{8}{3} dy = \left[ \frac{8}{3}y \right]_{0}^{4} = \frac{32}{3}\).

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Most popular questions from this chapter

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{0}^{1} \int_{0}^{2} d y d x $$

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