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Complete Example 6by evaluating the iterated integral -22-4-x24-x20x+z+4kydydzdx.

Short Answer

Expert verified

The value of the iterated integral -22-4-x24-x20x+z+4kydydzdxis,110k.

Step by step solution

01

Step 1. Given information

-22-4-x24-x20x+z+4kydydzdx.

02

Step 2. Evaluate the given triple integral.

-22-4-x24-x20x+z+4kydydzdx=-22-4-x24-x20x+z+4kydydzdx=-22-4-x24-x2ky220x+z+4dzdx=k2-22-4-x24-x2x+z+42dzdx

Now expand the obtained expression.

-22-4-x24-x20x+z+4kydydzdx=k2-22-4-x24-x2x2+z2+16+2xz+8z+8xdzdx

03

Step 3. Now integrate the obtained expression with respect to z.

-22-4-x24-x20x+z+4kydydzdx=k2-22x2z-4-x24-x2+z33-4-x24-x2+16z-4-x24-x2+2xz22-4-x24-x2+8z22-4-x24-x2+8xz-4-x24-x2=k2-22x224-x2+234-x23+1624-x2+x0+40+8x24-x2dx=k2-222x24-x2+234-x232+324-x2+16x4-x2dx

04

Step 4. Now integrate with respect to x.

-22-4-x24-x20x+z+4kydydzdx=k22-22x24-x2dx+23-224-x232dx+32-224-x2dx+16-22x4-x2dx=k22π+236π+322π+0=k22π+4π+64π=35=110k

Therefore, the required value is,110k.

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